PHYS 305 - INTRODUCTION TO MODERN PHYSICS - Simple harmonic motion Question Bank - Set 1 Liberty University Question 1 Question A block of mass m is attached to a horizontal spring with spring constant k. Initially, the block is at rest at the equilibrium position of the spring. At t = 0, a constant force F is applied to the block in the positive x-direction. Determine the amplitude of the resulting simple harmonic motion. Solution To determine the amplitude of the resulting simple harmonic motion, we need to consider the equilibrium position of the block and how it is displaced when the force F is applied. Step 1: At the equilibrium position, the net force on the block is zero. This means the force due to the spring is equal in magnitude and opposite in direction to the applied force: kx = F where x is the displacement of the block from the equilibrium position. Step 2: The block will accelerate in the positive x-direction until it reaches the maximum displacement, or the amplitude, denoted by A. At this point, the restoring force due to the spring will be equal in magnitude to the applied force F , but in the opposite direction: kA = F Step 3: To find the amplitude A, we can solve for it in terms of the given information: F A= k Therefore, the amplitude of the resulting simple harmonic motion is Fk . Question 2 Question A spring-mass system is experiencing simple harmonic motion with an amplitude 2 of 0.2 m and a period of 0.5 s. If the maximum acceleration of the mass is 8 m/s , determine the mass of the object. Solution Let’s denote the mass of the object as m, the angular frequency as ω, and the spring constant as k. Step 1: Recall the general form of simple harmonic motion. The displacement (x) of an object in simple harmonic motion can be described by the equation: x(t) = A sin(ωt + ϕ) where: - A is the amplitude, - ω is the angular frequency, - t is the time, and ϕ is the phase angle. Given that the amplitude is 0.2 m, the period is 0.5 s, and the maximum 2 acceleration is 8 m/s , we can relate these quantities to find the mass of the object. Step 2: Determine the angular frequency ω using the period T : T = 2π ω 0.5 = 2π ω 4π = 8π 0.5 Step 3: Recall the relationship between angular frequency ω, maximum acceleration Amax , and amplitude A: ω= Amax = ω 2 A 2 Given Amax = 8 m/s and A = 0.2 m, 8 = (8π)2 × 0.2 8 64π 2 m ≈ 0.012 kg m= Therefore, the mass of the object in the spring-mass system is approximately 0.012 kg. 2 Question 3 Question A 0.5 kg object is attached to a vertical spring with a spring constant of 200 N/m. The object is pulled down 0.1 m from its equilibrium position and released from rest. Find the amplitude, period, frequency, and maximum speed of the object during the motion. Solution Step 1: Find the amplitude of the oscillation. Amplitude = maximum displacement from equilibrium = 0.1 m Step 2: Find the period of the oscillation. r m Period = 2π k r 0.5 = 2π 200 √ ≈ 2π 0.0025 = 2π(0.05) = 0.314 s Step 3: Find the frequency of the oscillation. 1 Period 1 = 0.314 ≈ 3.18 Hz Frequency = Step 4: Find the maximum speed of the object. Maximum speed = Amplitude × angular frequency = 0.1 × 2π(3.18) ≈ 0.63 m/s Question 4 Question A mass-spring system oscillates with an amplitude of 0.2 m and a period of 2 seconds. If the maximum speed of the mass is 4 m/s, determine the displacement of the mass at this maximum speed. 3 Solution Step 1: Find the angular frequency of the system using the period. Given that the period T = 2 seconds and the angular frequency ω = 2π T . ω= 2π = π rad/s 2 Step 2: Use the amplitude and angular frequency to express the velocity. The √ velocity of an object undergoing simple harmonic motion is given by v(t) = ω A2 − x2 , where A is the amplitude and x is the displacement of the mass at time t. Given that the amplitude A = 0.2 m and the maximum velocity vmax = 4 m/s, substitute these values into the equation. p 4 = π 0.22 − x2 Step 3: Solve for the displacement x at the point of maximum velocity. Square both sides of the equation above and solve for x. 16 = π 2 (0.04 − x2 ) 16 = 0.04π 2 − π 2 x2 π 2 x2 = 0.04π 2 − 16 0.04π 2 − 16 π2 0.04 − 16 x2 = π2 −15.96 x2 = π2 x ≈ ±1.594 m x2 = Therefore, the displacement of the mass at the maximum speed is approximately 1.594 m. Question 5 Question A particle undergoes simple harmonic motion along the x-axis with an amplitude of 4 m and a frequency of 2 Hz. If the particle is at its equilibrium position at t = 0 s, determine the displacement and velocity of the particle at t = 1 s. 4 Solution Step 1: Find the expression for the displacement of the particle at any time t. The equation for simple harmonic motion along the x-axis is given by: x(t) = A cos(ωt + ϕ) where: - A is the amplitude, - ω = 2πf is the angular frequency, - ϕ is the phase angle. Given that A = 4 m and f = 2 Hz, we have: x(t) = 4 cos(4πt + ϕ) Step 2: Use the initial conditions to determine the phase angle ϕ. Since the particle is at its equilibrium position at t = 0 s, we have: x(0) = 4 cos(ϕ) = 0 This implies that ϕ = (since cos π2 = 0). Therefore, the equation for the displacement becomes: π x(t) = 4 cos 4πt + 2 Step 3: Calculate the displacement at t = 1 s. Substitute t = 1 s into the equation for displacement: π x(1) = 4 cos 4π + 2 9π x(1) = 4 cos = −4 m 2 Step 4: Find the velocity of the particle at t = 1 s. The velocity of the particle is given by: v(t) = −Aω sin(ωt + ϕ) π 2 Substitute the known values to get: π v(1) = −4 × 2π sin 4π + 2 9π v(1) = −8π sin = 0 m/s 2 Therefore, at t = 1 s, the displacement of the particle is -4 m and its velocity is 0 m/s. Question 6 Question A particle is in simple harmonic motion on a straight line. The amplitude of the motion is 4 cm and the period is 2 seconds. If the displacement of the particle at time t = 0.5 s is 2 cm in the positive direction, find the displacement of the particle at t = 1.5 s. 5 Solution Step 1: First, we write down the equation for simple harmonic motion: 2π x(t) = A cos t T where A is the amplitude and T is the period. Step 2: Given that the amplitude A = 4 cm and the period T = 2 s, the equation becomes: 2π x(t) = 4 cos t = 4 cos(πt) 2 Step 3: We are given that the displacement of the particle at time t = 0.5 s is 2 cm, so we can write: x(0.5) = 2 Substitute t = 0.5 into the equation and solve for the phase constant: π 4 cos =2 2 4·0=2 This implies that the phase constant is 0. Step 4: Now we can determine the displacement of the particle at t = 1.5 s: 3π = 4 · 0 = 0 cm x(1.5) = 4 cos(π · 1.5) = 4 cos 2 Therefore, the displacement of the particle at t = 1.5 s is 0 cm. Question 7 Question A particle of mass m is attached to a horizontal spring with spring constant k. Initially, the particle is at its equilibrium position. At time t = 0, the particle is displaced a distance A to the right and released from rest. Find the expression for the velocity of the particle as a function of time. Solution Step 1: Determine the angular frequency q ω k . The angular frequency ω is given by ω = m Step 2: Determine the amplitude of the motion The amplitude A is given in the problem as the maximum displacement of the particle. 6 Step 3: Write down the equation for the displacement x(t) The displacement of the particle as a function of time t can be described by the equation x(t) = A cos(ωt). Step 4: Find the velocity of the particle The velocity of the particle can be found by taking the derivative of the displacement function with respect to time: v(t) = dx = −Aω sin(ωt) dt Therefore, the expression for the velocity of the particle as a function of time is v(t) = −Aω sin(ωt). Question 8 Question A particle undergoes simple harmonic motion with an amplitude of 10 cm and a period of 4 seconds. If the particle is at its maximum displacement of 10 cm at time t = 0, determine a cosine function that describes the displacement of the particle as a function of time. Solution Step 1: First, let’s recall the general form of a cosine function for simple harmonic motion: x(t) = A cos(ωt + ϕ) where: - A is the amplitude, - ω is the angular frequency, given by 2π divided by the period (T ), so ω = 2π T , - ϕ is the phase angle. Step 2: Given that the amplitude A is 10 cm and the period T is 4 seconds, we can calculate the angular frequency ω: ω= 2π π = rad/s 4 2 Step 3: Since the particle is at its maximum displacement of 10 cm at time t = 0, we have: x(0) = 10 = 10 cos(ϕ) cos(ϕ) = 1 ϕ = 0 (since cos(0) = 1) Step 4: Therefore, the cosine function that describes the displacement of the particle as a function of time is: x(t) = 10 cos 7 π t 2 Question 9 Question A particle undergoes simple harmonic motion with a period of 4 seconds and an amplitude of 5 cm. Find the maximum velocity of the particle and the displacement when the velocity is half of the maximum. Solution Let’s denote the amplitude of the harmonic motion as A = 5 cm, the period as T = 4 seconds, the maximum velocity as vmax , and the displacement when the velocity is half of the maximum as xhalf . Step 1: Find the angular frequency The angular frequency ω of the simple harmonic motion is given by ω = 2π T . Substituting T = 4 seconds, we have: π 2π = rad/s ω= 4 2 Step 2: Find the maximum velocity The maximum velocity vmax of a particle undergoing simple harmonic motion is given by vmax = Aω. Substituting A = 5 cm and ω = π2 rad/s, we get: π 5π = cm/s 2 2 Step 3: Find the displacement when the velocity is half of the maximum When the velocity is half of the maximum, we have v = vmax = 5π 2 4 cm/s. The displacement x at this point can be found using the equation v = √ π ω A2 − x2 . Substitute v = 5π 4 , ω = 2 , and A = 5 into the equation: 5π πq 2 = 5 − x2half 4 2 Solving for xhalf gives: s r √ 2 r 5 25 5 3 75 2 = 25 − = = cm xhalf = 5 − 2 4 4 2 vmax = 5 × Therefore, the maximum velocity of the particle is 5π cm/s and the displace√ 2 5 3 ment when the velocity is half of the maximum is 2 cm. Question 10 Question A particle executes simple harmonic motion with an amplitude of 5 cm and a period of 4 seconds. If at time t = 0, the particle is at its equilibrium position and moving in the positive direction, determine the displacement of the particle at t = 2 seconds. 8 Solution Step 1: Calculate the angular frequency ω using the formula ω = 2π T , where T is the period. 2π π ω= = radians/second 4 2 Step 2: Write the displacement equation for simple harmonic motion: x(t) = A cos(ωt + ϕ) where A is the amplitude (5 cm) and ϕ is the phase angle. Step 3: Since at t = 0 the particle is at its equilibrium position and moving in the positive direction, the equation becomes: x(0) = A cos(ϕ) = 0 ⇒ cos(ϕ) = 0 This means that ϕ = π2 or 3π 2 . Step 4: Plug in ω, A, ϕ, and t = 2 seconds into the displacement equation to find the displacement at t = 2 seconds. π π π 3π x(2) = 5 cos ×2+ = −5 cm = 5 cos π + = 5 cos 2 2 2 2 Therefore, the displacement of the particle at t = 2 seconds is -5 cm. Question 11 Question A particle is in simple harmonic motion with an amplitude of 8 cm and a period of 2 seconds. If the particle is at its equilibrium position at time t = 0, find the equation describing the particle’s displacement as a function of time. Solution Step 1: Find the angular frequency ω using the period T . The angular frequency ω is given by: 2π ω= T Substitute T = 2 seconds into the equation: ω= 2π =π 2 Step 2: Determine the equation for the displacement of the particle. The equation for simple harmonic motion is given by: x(t) = A cos(ωt + ϕ) 9 where A is the amplitude, ω is the angular frequency, ϕ is the phase angle. Given that A = 8 cm, ω = π, and the particle is at its equilibrium position at t = 0: x(0) = 8 cos(ϕ) = 0 As the initial position is at the equilibrium position, cos(ϕ) = 0 which implies ϕ = π2 . Therefore, the equation describing the particle’s displacement as a function of time t is: π x(t) = 8 cos πt + 2 Question 12 Question A mass-spring system has a mass of 0.5 kg attached to a spring with a spring constant of 100 N/m. Initially, the mass is displaced 0.1 m from the equilibrium position and released from rest. Find the amplitude, angular frequency, and period of the resulting simple harmonic motion. Solution Step 1: Find the amplitude. The amplitude (A) of the simple harmonic motion is the maximum displacement of the mass from the equilibrium position. We are given that the mass is initially displaced 0.1 m from the equilibrium position, so A = 0.1 m. Step 2: Find the angular frequency. The angular frequency (ω) of the simple q k harmonic motion can be found using the formula ω = m , where k is the spring constant and m is the mass. Substituting in the given values, we have: r √ 100 √ ω= = 200 = 10 2 rad/s 0.5 Step 3: Find the period. The period (T ) of the simple harmonic motion is the time taken for one complete oscillation and is related to the angular frequency by T = 2π ω . Substituting in the value of ω we found in Step 2, we get: T = 2π π √ = √ ≈ 0.986 s 10 2 5 2 Therefore, the √ amplitude of the simple harmonic motion is 0.1 m, the angular frequency is 10 2 rad/s, and the period is approximately 0.986 s. Question 13 Question A particle performs simple harmonic motion with an amplitude of 4 cm and a period of 2 seconds. If the particle is at a distance of 2 cm from the mean 10 position at time t = 1 second, find an equation for the displacement of the particle at time t. Solution Step 1: Find the angular frequency ω. Given that the period T = 2 seconds, we 2π have ω = 2π T . Therefore, ω = 2 = π rad/s. Step 2: Find the equation for displacement. The general equation for simple harmonic motion is: x(t) = A cos(ωt + ϕ) where A is the amplitude, ω is the angular frequency, and ϕ is the phase constant. Step 3: Determine the phase constant ϕ. At time t = 1 second, the particle is 2 cm from the mean position. Thus, we have x(1) = 4 cos(π + ϕ) = 2. Solving for ϕ: 4 cos(π + ϕ) = 2 cos(π + ϕ) = π+ϕ= 1 2 π 3 2π π −π =− 3 3 Step 4: Substitute the values into the general equation. Therefore, the equation for the displacement of the particle at time t is: 2π x(t) = 4 cos πt − 3 ϕ= Thus, the displacement of the particle at any time t is given by x(t) = 4 cos πt − 2π 3 . Question 14 Question A block of mass m is attached to a horizontal spring with spring constant k. The block is displaced a distance A from the equilibrium position and released from rest. Find the maximum speed of the block during its oscillation. Solution Let’s denote the position of the block at time t as x(t). The equation of motion for simple harmonic motion is given by: m d2 x = −kx dt2 11 Given that the block is displaced a distance A from the equilibrium position, the initial conditions are x(0) = A and dx dt (0) = 0. Step 1: Solve the differential equation to find x(t). The solution to the differential equation is of the form: x(t) = Acos(ωt + ϕ) q k is the angular frequency and ϕ is the phase angle. where ω = m Step 2: Apply the initial condition x(0) = A. A = Acos(ϕ) Since cos(ϕ) = 1, we have ϕ = 0. Thus, the position function becomes: x(t) = Acos(ωt) Step 3: Find the maximum speed of the block. The speed of the block at any time t is given by: v(t) = −Aωsin(ωt) The maximum speed of the block is when sin(ωt) takes its maximum value of 1. Thus, the maximum speed vmax is: vmax = −Aω Substitute ω = q k m to get: r vmax = −A k m Question 15 Question A particle undergoes simple harmonic motion with an amplitude of 5 cm and a period of 1.5 seconds. If the displacement of the particle is 2.5 cm at time t = 0, determine: 1. The equation of motion for the particle. 2. The velocity of the particle at time t = 0.75 seconds. 3. The acceleration of the particle when it is 3 cm from the equilibrium position. 12 Solution 1. To find the equation of motion for the particle, we start with the general equation for simple harmonic motion: x(t) = A sin(ωt + ϕ) where: - A is the amplitude, - ω is the angular frequency, - ϕ is the phase angle. Given: Amplitude, A = 5 cm, Period, T = 1.5 s. We know that T = 2π ω , so we can solve for ω: 1.5 = 2π ω 2π 4π = 1.5 3 Now, we need to find the phase constant ϕ using the initial condition: At time t = 0, x(0) = 2.5 cm. Substituting this into the equation of motion: ω= x(0) = A sin(ϕ) = 2.5 5 sin(ϕ) = 2.5 2.5 = 0.5 5 π ϕ = sin−1 (0.5) = 6 Therefore, the equation of motion for the particle is: 4π π x(t) = 5 sin t+ 3 6 sin(ϕ) = 2. The velocity of the particle is given by the derivative of the displacement function: 4π π dx 4π =5 t+ v(t) = cos dt 3 3 6 At time t = 0.75 seconds: 4π 4π π v(0.75) = 5 cos × 0.75 + 3 3 6 4π π v(0.75) = 5 cos 2π + 3 6 4π 13π v(0.75) = 5 cos 3 6 √ ! 4π 3 v(0.75) = 5 − 3 2 13 √ v(0.75) = −10π 3 cm/s √ Therefore, the velocity of the particle at t = 0.75 seconds is −10π 3 cm/s. 3. The acceleration of the particle is given by the second derivative of the displacement function: 2 4π 4π π d2 x sin t+ a(t) = 2 = −5 dt 3 3 6 When the particle is 3 cm from the equilibrium position, we have: 4π π x(t) = 5 sin t+ =3 3 6 π 3 4π t+ = sin 3 6 5 To find the acceleration when x = 3 cm, we substitute x = 3 into the acceleration equation: 2 4π a(t) = −5 sin 3 Question 16 Question A particle undergoes simple harmonic motion with an amplitude of 3 cm and a period of 2 seconds. If the displacement of the particle is 2 cm at time t = 0, find the displacement at time t = 1 second. Solution Given that the particle undergoes simple harmonic motion, we can express its displacement x as a function of time t using the equation: x(t) = A cos(ωt + ϕ) where A is the amplitude, ω is the angular frequency, and ϕ is the phase angle. Step 1: Find the angular frequency ω using the period: Since T = 2π ω , we 2π have ω = 2π = = π rad/s. T 2 Step 2: Determine the phase angle ϕ: Given that the displacement of the particle is 2 cm at time t = 0, we have: x(0) = 3 cos ϕ = 2 Solving this equation for ϕ, we find ϕ = cos−1 23 . Step 3: Express the displacement x as a function of time t: Putting everything together, the displacement function becomes: 2 −1 x(t) = 3 cos πt + cos 3 14 Step 4: Find the particle’s displacement at time t = 1 second: Substitute t = 1 into the displacement function: 2 −1 x(1) = 3 cos π + cos 3 Since cos(π + θ) = − cos(θ), the displacement simplifies to: 2 2 −1 x(1) = −3 cos cos = −3 = −2 cm 3 3 Therefore, the displacement of the particle at time t = 1 second is −2 cm. Question 17 Question A particle undergoes simple harmonic motion with an amplitude of 5 cm and a period of 2 seconds. If the particle is located at 3 cm at time t = 0, find the displacement function s(t) for the particle. Solution Step 1: Identify the parameters of the simple harmonic motion. Given: Amplitude, A = 5 cm, Period, T = 2 seconds, Initial displacement, s(0) = 3 cm, Displacement function, s(t) = A sin 2π T t + ϕ , where ϕ is the phase angle. Step 2: Determine the angular frequency. The angular frequency, ω, can be 2π calculated using the formula ω = 2π T . ω = 2 = π radians/second. Step 3: Find the phase angle, ϕ. Given that s(0) = 3 cm, we substitute t = 0 in the displacement function to find ϕ. s(0) = 5sin(ϕ) = 3. This implies that sin(ϕ) = 53 . From this, we find that ϕ = sin−1 35 . Step 4: Write the displacement function s(t). Therefore, the displacement function is s(t) = 5 sin πt + sin−1 53 . Question 18 Question A particle undergoes simple harmonic motion with an amplitude of 5 cm and a period of 2 seconds. If the maximum velocity of the particle is 10 m/s, determine the equation of motion for the particle. Solution Step 1: Identify the given values and parameters of the simple harmonic motion. Given: Amplitude A = 5 cm = 0.05 m Period T = 2 s Maximum velocity vmax = 10 m/s 15 Step 2: Find the angular frequency ω. The angular frequency can be calculated using the formula: 2π ω= T 2π ω= = π rad/s 2 Step 3: Determine the equation of motion for the simple harmonic motion. The general equation for simple harmonic motion is given by: x(t) = A cos(ωt + ϕ) where x(t) is the displacement at time t, A is the amplitude, ω is the angular frequency, and ϕ is the phase angle. Step 4: Find the phase angle ϕ. Since the maximum velocity occurs when the displacement is zero, at this point the particle is at the equilibrium position and the cosine term is at its maximum value of 1. Therefore, ϕ is 0. Step 5: Substitute the values into the equation of motion. Substitute A = 0.05 m, ω = π rad/s, and ϕ = 0 into the equation of motion to get: x(t) = 0.05 cos(πt) Therefore, the equation of motion for the particle undergoing simple harmonic motion is x(t) = 0.05 cos(πt). Question 19 Question A particle of mass m is attached to a spring with spring constant k. Initially, the particle is at rest at the equilibrium position. At time t = 0, the particle is given an initial velocity v0 in the positive x-direction. Determine the maximum elongation of the spring during the subsequent motion. Solution Let’s denote the equilibrium position of the particle as x = 0. At any time t, the elongation of the spring from the equilibrium position is given by x(t). Step 1: Write the equation of motion for the particle. The equation of motion for simple harmonic motion is given by m d2 x = −kx dt2 Given that the particle is given an initial velocity v0 in the positive x-direction at t = 0, we have the initial conditions x(0) = 0 and 16 dx dt = v0 t=0 Step 2: Solve the differential equation. To solve this second-order differential equation, we can guess a solution of the form x(t) = A cos(ωt) + B sin(ωt) and find ω using k/m = ω 2 . Differentiating x(t) twice with respect to t and substiting back into the differential equation will give us the values of A and B. Step 3: Find the maximum elongation of the spring. The maximum elongation of the spring occurs at the amplitude of the oscillation. In this case, the amplitude is the maximum positive value of x(t), which can be found once A and B are determined. After solving for A and B, we can write down the equation for x(t) and then determine the maximum elongation from the amplitude of the resulting oscillatory function. Question 20 Question A mass m is attached to a spring with spring constant k. The mass is displaced from its equilibrium position and released. If the time period of the resulting simple harmonic motion is 5 seconds, find the frequency of the oscillation in Hz. Solution Step 1: Recall that the frequency (f ) of an oscillation is the reciprocal of the time period (T ), i.e., f = T1 . Step 2: Given that the time period T = 5 seconds, we can find the frequency f by substituting T into the formula: f = 15 Hz. Step 3: Therefore, the frequency of the oscillation of the mass attached to the spring is 15 Hz. Question 21 Question A particle undergoes simple harmonic motion with an amplitude of 0.2 m and a frequency of 2 Hz. If the displacement of the particle is given by x = 0.1 cos(4πt), where x is in meters and t is in seconds, determine the phase angle, period, angular frequency, and maximum velocity of the particle. Solution Step 1: To determine the phase angle, we compare the equation of motion with the general equation of simple harmonic motion. The general equation is given by x = A cos(ωt + ϕ), where A is the amplitude, ω is the angular frequency, and ϕ is the phase angle. By comparing the two equations, we see that ϕ = 0. 17 Step 2: The period of the motion can be found using the relation T = f1 , where T is the period and f is the frequency. Substituting the values given, we get T = 12 = 0.5 s. Step 3: The angular frequency is related to the frequency by the equation ω = 2πf . Substituting the frequency, we find ω = 2π × 2 = 4π rad/s. Step 4: The maximum velocity of the particle occurs when the displacement is zero. The velocity of a particle undergoing simple harmonic motion is given by v = −Aω sin(ωt + ϕ). At the equilibrium position (x = 0), the velocity is maximized. Substituting the values A = 0.1, ω = 4π, ϕ = 0, we find that the maximum velocity is 0.4π m/s. Question 22 Question A particle of mass 0.5 kg is attached to a spring with a spring constant of 20 N/m. It is initially displaced 0.1 m from its equilibrium position and released from rest. Find an expression for the position of the particle as a function of time. Solution Step 1: From Hooke’s Law, the force acting on the particle at any time t is given by F = −kx, where k is the spring constant and x is the displacement from equilibrium. Step 2: According to Newton’s Second Law, the acceleration of the particle F is given by a = , where m is the mass of the particle. m Step 3: Substituting F = −kx into the equation for acceleration, we have k a = − x. m Step 4: The acceleration can also be expressed as the second derivative of d2 x displacement with respect to time, i.e., a = 2 . dt k d2 x Step 5: Equating the two expressions for acceleration, we get 2 = − x. dt m Step 6: This is a second-order linear differential equation with constant coefficients, the general solution of which is x(t) =rA cos(ωt) + B sin(ωt), where k A and B are constants to be determined and ω = is the angular frequency. m Step 7: Applying the initial conditions x(0) = 0.1 and v(0) = 0, where dx v(t) = is the velocity of the particle, we can find the values of A and B. dt Step 8: After finding A and B, the final expression for the position of the particle as a function of time is x(t) = A cos(ωt) + B sin(ωt). 18 Question 23 Question A particle undergoes simple harmonic motion with an amplitude of 3 cm and a frequency of 2 Hz. If the particle is at a displacement of 2 cm at time t = 0, determine the equation for the displacement of the particle as a function of time. Solution Given parameters: Amplitude, A = 3 cm Frequency, f = 2 Hz Displacement at t = 0, s(0) = 2 cm The general form of the displacement equation for simple harmonic motion is: s(t) = A sin(ωt + ϕ) where ω = 2πf is the angular frequency and ϕ is the phase angle. Step 1: Find the angular frequency, ω ω = 2πf = 2π × 2 = 4π rad/s Step 2: Find the phase angle, ϕ Given s(0) = 2 cm: s(0) = A sin(ϕ) = 3 × sin(ϕ) = 2 Solving for ϕ: 2 3 2 −1 ⇒ ϕ = sin 3 sin(ϕ) = Step 3: Write the equation of motion The equation for the displacement of the particle as a function of time is: s(t) = 3 sin 4πt + sin−1 (2/3) Question 24 Question A mass-spring system is oscillating with an amplitude of 0.2 m. If the maximum velocity of the mass is 0.5 m/s, determine the frequency of the oscillation. 19 Solution Step 1: We know that the maximum velocity of a mass-spring system in simple harmonic motion is given by vmax = Aω, where A is the amplitude and ω is the angular frequency. Step 2: Given that vmax = 0.5 m/s and A = 0.2 m, we can write 0.5 = 0.2ω. Step 3: Solving for ω, we find ω = 0.5 0.2 = 2.5 rad/s. Step 4: The frequency f of simple harmonic motion is related to the angular frequency ω ω by the formula f = 2π . Step 5: Substituting ω = 2.5 rad/s into the formula 2.5 gives f = 2π ≈ 0.3979 Hz. Step 6: Therefore, the frequency of the oscillation is approximately 0.3979 Hz. Question 25 Question A block of mass m is attached to a spring with spring constant k. The block is pulled to a distance A from the equilibrium position and released. Determine the maximum speed of the block during the oscillation. Solution Let’s denote the equilibrium position of the block as x = 0, the initial position as x = A, and the maximum displacement from equilibrium as x = ±A. The potential energy of the block-spring system at any position x is given by P E = 1 2 2 kx . Step 1: The total mechanical energy E of the system is conserved and is the sum of kinetic and potential energies. E = KE + P E Initially, when the block is at position x = A, all energy is potential energy, hence E = P E(A) = 21 kA2 . Step 2: When the block reaches the equilibrium position at x = 0, all 2 energy is kinetic energy. Therefore, E = KEmax = 21 mvmax . Step 3: Since energy is conserved, the total energy at x = A must equal the total energy at x = 0. 1 2 1 2 kA = mvmax 2 2 Step 4: Solving for the maximum speed vmax of the block: r k 2 vmax = A m q k 2 A . Hence, the maximum speed of the block during oscillation is vmax = m 20 Question 26 Question A mass-spring system has a mass of 0.5 kg attached to a spring with a spring constant of 20 N/m. If the maximum velocity of the mass is 1 m/s, find the amplitude of the oscillation. Solution Step 1: Find the angular frequency of the oscillation using the formula: r k ω= m where ω is the angular frequency, k is the spring constant, and m is the mass. r √ √ 20 ω= = 40 = 2 10 ≈ 6.32 rad/s 0.5 Step 2: The maximum velocity in simple harmonic motion is given by the formula: vmax = ωA where vmax is the maximum velocity and A is the amplitude of the oscillation. Given vmax = 1 m/s, we can rearrange the formula to solve for A: A= vmax 1 1√ = √ = 10 ≈ 1.58 m ω 2 2 10 Therefore, the amplitude of the oscillation is approximately 1.58 m. Question 27 Question A mass m is attached to a spring with spring constant k. The mass is released from rest at its equilibrium position. It reaches its maximum displacement of A from the equilibrium position, then returns to its equilibrium position before continuing to the maximum displacement of A in the opposite direction. Determine the period of oscillation of the mass. Solution To determine the period of oscillation, we first need to find the angular frequency ω, and then we can use the formula for the period T = 2π ω . Step 1: Find the angular frequency ω. The angular frequency ω can be found using the formula: r k ω= m 21 Step 2: Find the period T . The period of oscillation, T , is given by: T = Since ω = q k m, 2π ω we have: 2π T = q = 2π k m r m k Therefore, the period of oscillation of the mass is 2π pm k . Question 28 Question A particle of mass 0.5 kg is attached to a spring with a spring constant of 20 N/m. Initially, the particle is displaced 0.1 m from its equilibrium position and released from rest. Determine the amplitude, frequency, angular frequency, and period of the resulting simple harmonic motion. Solution Step 1: Find the amplitude A A = 0.1 m Step 2: Find the angular frequency ω r r √ k 20 ω= = = 40 rad/s m 0.5 Step 3: Find the frequency f √ ω 40 6.32 f= = ≈ Hz 2π 2π 2π Step 4: Find the period T T = 1 1 1 = ≈ s ≈ 0.995 s f 6.32/2π 1.005 Therefore,√the amplitude of the simple harmonic motion is 0.1 m, the angular frequency is 40 rad/s, the frequency is approximately 6.32 2π Hz, and the period is approximately 0.995 s. 22 Question 29 Question A particle undergoes simple harmonic motion with an amplitude of 5 cm and a frequency of 2 Hz. If the particle is at its equilibrium position at time t = 0, determine the displacement, velocity, and acceleration of the particle at time t = 32 seconds. Solution Step 1: Find the angular frequency ω The angular frequency ω is related to the frequency f by the equation ω = 2πf . Given that f = 2 Hz, we can compute ω by ω = 2π × 2 = 4π rad/s Step 2: Determine the displacement at t = 23 s The displacement at time t for a particle undergoing simple harmonic motion is given by x(t) = A sin(ωt + ϕ) where A is the amplitude, ω is the angular frequency, t is time, and ϕ is the phase angle. Since the particle is at its equilibrium position at t = 0, we have ϕ = 0. Thus, the displacement at time t = 32 s is 2 2 8π x = 5 sin 4π × = 5 sin 3 3 3 √ √ 2π 3 5 3 2 = 5 sin =5× = cm x 3 3 2 2 Step 3: Calculate the velocity at t = The velocity of the particle is given by 2 3 s v(t) = Aω cos(ωt + ϕ) Substitute A = 5, ω = 4π, t = 23 , and ϕ = 0: 2 2 8π v = 5 × 4π cos 4π × = 20π cos 3 3 3 2 2π 1 v = 20π cos = −20π × = −10π cm/s 3 3 2 Step 4: Find the acceleration at t = 32 s The acceleration of the particle is given by a(t) = −Aω 2 sin(ωt + ϕ) 23 Substitute A = 5, ω = 4π, t = 23 , and ϕ = 0: 2 2 8π a = −5 × (4π)2 sin 4π × = −80π 2 sin 3 3 3 √ √ 3 2 2π 2 a = −80π 2 sin = −80π 2 × = −40π 2 3 cm/s 3 3 2 Therefore, at t = 23 s, the displacement is √ and acceleration is −40π 2 3 cm/s2 . √ 5 3 2 cm, velocity is −10π cm/s, Question 30 Question An object of mass m is attached to a spring with spring constant k. The object is displaced from its equilibrium position by a distance x0 and released from rest. Find the maximum speed of the object during its motion. Solution Step 1: Determine the maximum potential energy of the object. The maximum potential energy occurs when the object is at its maximum displacement x0 . The potential energy at this position is given by P E = 21 kx20 . Step 2: Use conservation of mechanical energy to find the maximum speed. At the equilibrium position, all the potential energy is converted to kinetic energy. Therefore, the maximum potential energy at x0 equals the maximum kinetic energy at the equilibrium position. The kinetic energy is given by KE = 1 2 2 mvmax , where vmax is the maximum speed of the object. Step 3: Equate the potential energy and kinetic energy expressions. 2 We have 12 kx20 = 12 mvmax . Step 4: Solve for the maximum speed vmax . q q x0 k 2 k From the equation in Step 3, we find vmax = m x0 = x0 m . Therefore, the maximum speed of the object during its motion is vmax = q k m. Question 31 Question A mass attached to a spring executes simple harmonic motion with an amplitude of 5 cm and a period of 2 seconds. If the maximum acceleration of the mass is 2 8 m/s , determine the mass and the spring constant. 24 Solution Step 1: Identify the formula for the period of simple harmonic motion: The period (T ) of simple harmonic motion is related to the angular frequency (ω) by the equation: 2π T = ω Step 2: Determine the angular frequency from the period: Given that the period (T ) is 2 seconds, we can find the angular frequency (ω) using the equation from Step 1: 2π 2π = = π rad/s ω= T 2 Step 3: Determine the maximum velocity from the amplitude and period: The maximum velocity in simple harmonic motion is given by: vmax = ω · amplitude Given that the amplitude is 5 cm (or 0.05 m), we can find the maximum velocity: vmax = π × 0.05 = 0.157 m/s Step 4: Determine the mass using the maximum acceleration: The acceleration in simple harmonic motion is related to the angular frequency and the amplitude by: amax = ω 2 · amplitude 2 Given that the maximum acceleration is 8 m/s , we can find the mass (m): m= amax 8 = 2 ≈ 0.81 kg ω2 π Step 5: Determine the spring constant using the mass and maximum acceleration: The spring constant (k) can be found using the equation: k= m · amax 0.81 × 8 = = 129.6 N/m amplitude 0.05 Therefore, the mass of the object is approximately 0.81 kg and the spring constant is approximately 129.6 N/m. Question 32 Question A 0.5 kg mass attached to a spring oscillates with a frequency of 2 Hz. If the amplitude of the oscillation is 0.1 m, what is the maximum speed of the mass during the oscillation? 25 Solution Step 1: We know that the frequency of oscillation, f , is related to the angular ω frequency, ω, by the formula f = 2π . Therefore, we can find ω using the given frequency. Step 1: f = 2 Hz ω = 2πf = 2π × 2 = 4π rad/s Step 2: The maximum speed of the mass during simple harmonic motion is given by the formula vmax = ωA, where A is the amplitude of oscillation. Step 2: A = 0.1 m vmax = 4π × 0.1 = 0.4π m/s Therefore, the maximum speed of the mass during the oscillation is 0.4π ≈ 1.26 m/s. Question 33 Question A spring-mass system is set into oscillation with an amplitude of 10 cm and a frequency of 2 Hz. If the maximum speed of the mass is 20 cm/s, determine the displacement of the mass when its speed is 10 cm/s. Solution Step 1: Determine the angular frequency (ω) using the formula f = ω 2π . ω = 2π × f = 2π × 2 = 4π rad/s Step 2: Calculate the maximum displacement of the mass using the formula A = Vmax ω , where A is the amplitude and Vmax is the maximum speed. 20 5 ⇒ 10 = rad 4π 2π Step 3: Determine the displacement when the speed is 10 cm/s. Let the displacement be x. 5t 5t ⇒ cos =1 10 = 10 cos(ωt) = 10 cos 2π 2π 10 = Step 4: Find the corresponding time t for speed 10 cm/s. 5t =0⇒t=0s 2π Step 5: Substitute the time into the equation for displacement. x = 10 cos(0) = 10 cm Therefore, the displacement of the mass when its speed is 10 cm/s is 10 cm. 26 Question 34 Question A particle undergoes simple harmonic motion with an amplitude of 5 cm and a period of 2 seconds. If the displacement of the particle is 3 cm at time t = 1 second, find the equation of motion for the particle. Solution Step 1: Determine the angular frequency ω 2π We know that the angular frequency ω = , where T is the period. Given T that T = 2 seconds, we have ω= 2π = π rad/s 2 Step 2: Determine the equation of motion The general equation of motion for simple harmonic motion is given by x(t) = A cos(ωt + ϕ), where: - A is the amplitude of the motion, - ω is the angular frequency, - t is time, and - ϕ is the phase angle. Given that the amplitude A = 5 cm and the particle is at 3 cm at t = 1 second, we can substitute these values into the general equation and solve for the phase angle ϕ: 3 = 5 cos(π + ϕ) Step 3: Solve for ϕ First, we need to find the quadrant in which the particle is located at t = 1 second. Since the particle is at 3 cm and the amplitude is 5 cm, the particle is at the negative end of its motion. Therefore, cos(π) = −1, and we have 3 = 5(−1) = −5 Solving for ϕ: −5 = 5 cos(ϕ) cos(ϕ) = −1 ϕ=π Step 4: Write the equation of motion Now we have found the phase angle ϕ = π. Substituting this into the general equation, we get: x(t) = 5 cos(πt + π) Therefore, the equation of motion for the particle is x(t) = 5 cos(πt + π). 27 Question 35 Question A particle is undergoing simple harmonic motion with an amplitude of 6 cm and a frequency of 2 Hz. If at t = 0 s the particle is at its maximum displacement from the equilibrium position of 3 cm in the positive direction, determine the equation of motion for the particle. Solution Step 1: Determine the angular frequency ω. Given that the frequency is 2 Hz, ω to find ω. we can use the relation f = 2π f = 2 Hz = ω 2π ω = 4π rad/s Step 2: Write the equation of motion for simple harmonic motion. The equation of motion for simple harmonic motion is given by: x(t) = A cos(ωt + ϕ) where: - x(t) is the displacement of the particle at time t, - A is the amplitude of the motion, - ω is the angular frequency, - ϕ is the phase angle. Step 3: Substitute the given values and determine the phase angle, ϕ. With the given information, we have: - Amplitude, A = 6 cm - Initial displacement, x(0) = 3 cm Plugging in these values to the general equation of motion: x(0) = A cos(ϕ) = 3 6 cos(ϕ) = 3 cos(ϕ) = 1 2 π rad 3 Step 4: Write the equation of motion for the particle. Finally, plugging in the known values into the equation of motion: π x(t) = 6 cos 4πt + 3 ϕ= 28 Question 2 Question A spring-mass system is experiencing simple harmonic motion with an amplitude 2 of 0.2 m and a period of 0.5 s. If the maximum acceleration of the mass is 8 m/s , determine the mass of the object. Solution Let’s denote the mass of the object as m, the angular frequency as ω, and the spring constant as k. Step 1: Recall the general form of simple harmonic motion. The displacement (x) of an object in simple harmonic motion can be described by the equation: x(t) = A sin(ωt + ϕ) where: - A is the amplitude, - ω is the angular frequency, - t is the time, and ϕ is the phase angle. Given that the amplitude is 0.2 m, the period is 0.5 s, and the maximum 2 acceleration is 8 m/s , we can relate these quantities to find the mass of the object. Step 2: Determine the angular frequency ω using the period T : T = 2π ω 0.5 = 2π ω 4π = 8π 0.5 Step 3: Recall the relationship between angular frequency ω, maximum acceleration Amax , and amplitude A: ω= Amax = ω 2 A 2 Given Amax = 8 m/s and A = 0.2 m, 8 = (8π)2 × 0.2 8 64π 2 m ≈ 0.012 kg m= Therefore, the mass of the object in the spring-mass system is approximately 0.012 kg. 2 Question 3 Question A 0.5 kg object is attached to a vertical spring with a spring constant of 200 N/m. The object is pulled down 0.1 m from its equilibrium position and released from rest. Find the amplitude, period, frequency, and maximum speed of the object during the motion. Solution Step 1: Find the amplitude of the oscillation. Amplitude = maximum displacement from equilibrium = 0.1 m Step 2: Find the period of the oscillation. r m Period = 2π k r 0.5 = 2π 200 √ ≈ 2π 0.0025 = 2π(0.05) = 0.314 s Step 3: Find the frequency of the oscillation. 1 Period 1 = 0.314 ≈ 3.18 Hz Frequency = Step 4: Find the maximum speed of the object. Maximum speed = Amplitude × angular frequency = 0.1 × 2π(3.18) ≈ 0.63 m/s Question 4 Question A mass-spring system oscillates with an amplitude of 0.2 m and a period of 2 seconds. If the maximum speed of the mass is 4 m/s, determine the displacement of the mass at this maximum speed. 3 Solution Step 1: Find the angular frequency of the system using the period. Given that the period T = 2 seconds and the angular frequency ω = 2π T . ω= 2π = π rad/s 2 Step 2: Use the amplitude and angular frequency to express the velocity. The √ velocity of an object undergoing simple harmonic motion is given by v(t) = ω A2 − x2 , where A is the amplitude and x is the displacement of the mass at time t. Given that the amplitude A = 0.2 m and the maximum velocity vmax = 4 m/s, substitute these values into the equation. p 4 = π 0.22 − x2 Step 3: Solve for the displacement x at the point of maximum velocity. Square both sides of the equation above and solve for x. 16 = π 2 (0.04 − x2 ) 16 = 0.04π 2 − π 2 x2 π 2 x2 = 0.04π 2 − 16 0.04π 2 − 16 π2 0.04 − 16 x2 = π2 −15.96 x2 = π2 x ≈ ±1.594 m x2 = Therefore, the displacement of the mass at the maximum speed is approximately 1.594 m. Question 5 Question A particle undergoes simple harmonic motion along the x-axis with an amplitude of 4 m and a frequency of 2 Hz. If the particle is at its equilibrium position at t = 0 s, determine the displacement and velocity of the particle at t = 1 s. 4 Solution Step 1: Find the expression for the displacement of the particle at any time t. The equation for simple harmonic motion along the x-axis is given by: x(t) = A cos(ωt + ϕ) where: - A is the amplitude, - ω = 2πf is the angular frequency, - ϕ is the phase angle. Given that A = 4 m and f = 2 Hz, we have: x(t) = 4 cos(4πt + ϕ) Step 2: Use the initial conditions to determine the phase angle ϕ. Since the particle is at its equilibrium position at t = 0 s, we have: x(0) = 4 cos(ϕ) = 0 This implies that ϕ = (since cos π2 = 0). Therefore, the equation for the displacement becomes: π x(t) = 4 cos 4πt + 2 Step 3: Calculate the displacement at t = 1 s. Substitute t = 1 s into the equation for displacement: π x(1) = 4 cos 4π + 2 9π x(1) = 4 cos = −4 m 2 Step 4: Find the velocity of the particle at t = 1 s. The velocity of the particle is given by: v(t) = −Aω sin(ωt + ϕ) π 2 Substitute the known values to get: π v(1) = −4 × 2π sin 4π + 2 9π v(1) = −8π sin = 0 m/s 2 Therefore, at t = 1 s, the displacement of the particle is -4 m and its velocity is 0 m/s. Question 6 Question A particle is in simple harmonic motion on a straight line. The amplitude of the motion is 4 cm and the period is 2 seconds. If the displacement of the particle at time t = 0.5 s is 2 cm in the positive direction, find the displacement of the particle at t = 1.5 s. 5 Solution Step 1: First, we write down the equation for simple harmonic motion: 2π x(t) = A cos t T where A is the amplitude and T is the period. Step 2: Given that the amplitude A = 4 cm and the period T = 2 s, the equation becomes: 2π x(t) = 4 cos t = 4 cos(πt) 2 Step 3: We are given that the displacement of the particle at time t = 0.5 s is 2 cm, so we can write: x(0.5) = 2 Substitute t = 0.5 into the equation and solve for the phase constant: π 4 cos =2 2 4·0=2 This implies that the phase constant is 0. Step 4: Now we can determine the displacement of the particle at t = 1.5 s: 3π = 4 · 0 = 0 cm x(1.5) = 4 cos(π · 1.5) = 4 cos 2 Therefore, the displacement of the particle at t = 1.5 s is 0 cm. Question 7 Question A particle of mass m is attached to a horizontal spring with spring constant k. Initially, the particle is at its equilibrium position. At time t = 0, the particle is displaced a distance A to the right and released from rest. Find the expression for the velocity of the particle as a function of time. Solution Step 1: Determine the angular frequency q ω k . The angular frequency ω is given by ω = m Step 2: Determine the amplitude of the motion The amplitude A is given in the problem as the maximum displacement of the particle. 6 Step 3: Write down the equation for the displacement x(t) The displacement of the particle as a function of time t can be described by the equation x(t) = A cos(ωt). Step 4: Find the velocity of the particle The velocity of the particle can be found by taking the derivative of the displacement function with respect to time: v(t) = dx = −Aω sin(ωt) dt Therefore, the expression for the velocity of the particle as a function of time is v(t) = −Aω sin(ωt). Question 8 Question A particle undergoes simple harmonic motion with an amplitude of 10 cm and a period of 4 seconds. If the particle is at its maximum displacement of 10 cm at time t = 0, determine a cosine function that describes the displacement of the particle as a function of time. Solution Step 1: First, let’s recall the general form of a cosine function for simple harmonic motion: x(t) = A cos(ωt + ϕ) where: - A is the amplitude, - ω is the angular frequency, given by 2π divided by the period (T ), so ω = 2π T , - ϕ is the phase angle. Step 2: Given that the amplitude A is 10 cm and the period T is 4 seconds, we can calculate the angular frequency ω: ω= 2π π = rad/s 4 2 Step 3: Since the particle is at its maximum displacement of 10 cm at time t = 0, we have: x(0) = 10 = 10 cos(ϕ) cos(ϕ) = 1 ϕ = 0 (since cos(0) = 1) Step 4: Therefore, the cosine function that describes the displacement of the particle as a function of time is: x(t) = 10 cos 7 π t 2 Question 9 Question A particle undergoes simple harmonic motion with a period of 4 seconds and an amplitude of 5 cm. Find the maximum velocity of the particle and the displacement when the velocity is half of the maximum. Solution Let’s denote the amplitude of the harmonic motion as A = 5 cm, the period as T = 4 seconds, the maximum velocity as vmax , and the displacement when the velocity is half of the maximum as xhalf . Step 1: Find the angular frequency The angular frequency ω of the simple harmonic motion is given by ω = 2π T . Substituting T = 4 seconds, we have: π 2π = rad/s ω= 4 2 Step 2: Find the maximum velocity The maximum velocity vmax of a particle undergoing simple harmonic motion is given by vmax = Aω. Substituting A = 5 cm and ω = π2 rad/s, we get: π 5π = cm/s 2 2 Step 3: Find the displacement when the velocity is half of the maximum When the velocity is half of the maximum, we have v = vmax = 5π 2 4 cm/s. The displacement x at this point can be found using the equation v = √ π ω A2 − x2 . Substitute v = 5π 4 , ω = 2 , and A = 5 into the equation: 5π πq 2 = 5 − x2half 4 2 Solving for xhalf gives: s r √ 2 r 5 25 5 3 75 2 = 25 − = = cm xhalf = 5 − 2 4 4 2 vmax = 5 × Therefore, the maximum velocity of the particle is 5π cm/s and the displace√ 2 5 3 ment when the velocity is half of the maximum is 2 cm. Question 10 Question A particle executes simple harmonic motion with an amplitude of 5 cm and a period of 4 seconds. If at time t = 0, the particle is at its equilibrium position and moving in the positive direction, determine the displacement of the particle at t = 2 seconds. 8 Solution Step 1: Calculate the angular frequency ω using the formula ω = 2π T , where T is the period. 2π π ω= = radians/second 4 2 Step 2: Write the displacement equation for simple harmonic motion: x(t) = A cos(ωt + ϕ) where A is the amplitude (5 cm) and ϕ is the phase angle. Step 3: Since at t = 0 the particle is at its equilibrium position and moving in the positive direction, the equation becomes: x(0) = A cos(ϕ) = 0 ⇒ cos(ϕ) = 0 This means that ϕ = π2 or 3π 2 . Step 4: Plug in ω, A, ϕ, and t = 2 seconds into the displacement equation to find the displacement at t = 2 seconds. π π π 3π x(2) = 5 cos ×2+ = −5 cm = 5 cos π + = 5 cos 2 2 2 2 Therefore, the displacement of the particle at t = 2 seconds is -5 cm. Question 11 Question A particle is in simple harmonic motion with an amplitude of 8 cm and a period of 2 seconds. If the particle is at its equilibrium position at time t = 0, find the equation describing the particle’s displacement as a function of time. Solution Step 1: Find the angular frequency ω using the period T . The angular frequency ω is given by: 2π ω= T Substitute T = 2 seconds into the equation: ω= 2π =π 2 Step 2: Determine the equation for the displacement of the particle. The equation for simple harmonic motion is given by: x(t) = A cos(ωt + ϕ) 9 where A is the amplitude, ω is the angular frequency, ϕ is the phase angle. Given that A = 8 cm, ω = π, and the particle is at its equilibrium position at t = 0: x(0) = 8 cos(ϕ) = 0 As the initial position is at the equilibrium position, cos(ϕ) = 0 which implies ϕ = π2 . Therefore, the equation describing the particle’s displacement as a function of time t is: π x(t) = 8 cos πt + 2 Question 12 Question A mass-spring system has a mass of 0.5 kg attached to a spring with a spring constant of 100 N/m. Initially, the mass is displaced 0.1 m from the equilibrium position and released from rest. Find the amplitude, angular frequency, and period of the resulting simple harmonic motion. Solution Step 1: Find the amplitude. The amplitude (A) of the simple harmonic motion is the maximum displacement of the mass from the equilibrium position. We are given that the mass is initially displaced 0.1 m from the equilibrium position, so A = 0.1 m. Step 2: Find the angular frequency. The angular frequency (ω) of the simple q k harmonic motion can be found using the formula ω = m , where k is the spring constant and m is the mass. Substituting in the given values, we have: r √ 100 √ ω= = 200 = 10 2 rad/s 0.5 Step 3: Find the period. The period (T ) of the simple harmonic motion is the time taken for one complete oscillation and is related to the angular frequency by T = 2π ω . Substituting in the value of ω we found in Step 2, we get: T = 2π π √ = √ ≈ 0.986 s 10 2 5 2 Therefore, the √ amplitude of the simple harmonic motion is 0.1 m, the angular frequency is 10 2 rad/s, and the period is approximately 0.986 s. Question 13 Question A particle performs simple harmonic motion with an amplitude of 4 cm and a period of 2 seconds. If the particle is at a distance of 2 cm from the mean 10 position at time t = 1 second, find an equation for the displacement of the particle at time t. Solution Step 1: Find the angular frequency ω. Given that the period T = 2 seconds, we 2π have ω = 2π T . Therefore, ω = 2 = π rad/s. Step 2: Find the equation for displacement. The general equation for simple harmonic motion is: x(t) = A cos(ωt + ϕ) where A is the amplitude, ω is the angular frequency, and ϕ is the phase constant. Step 3: Determine the phase constant ϕ. At time t = 1 second, the particle is 2 cm from the mean position. Thus, we have x(1) = 4 cos(π + ϕ) = 2. Solving for ϕ: 4 cos(π + ϕ) = 2 cos(π + ϕ) = π+ϕ= 1 2 π 3 2π π −π =− 3 3 Step 4: Substitute the values into the general equation. Therefore, the equation for the displacement of the particle at time t is: 2π x(t) = 4 cos πt − 3 ϕ= Thus, the displacement of the particle at any time t is given by x(t) = 4 cos πt − 2π 3 . Question 14 Question A block of mass m is attached to a horizontal spring with spring constant k. The block is displaced a distance A from the equilibrium position and released from rest. Find the maximum speed of the block during its oscillation. Solution Let’s denote the position of the block at time t as x(t). The equation of motion for simple harmonic motion is given by: m d2 x = −kx dt2 11 Given that the block is displaced a distance A from the equilibrium position, the initial conditions are x(0) = A and dx dt (0) = 0. Step 1: Solve the differential equation to find x(t). The solution to the differential equation is of the form: x(t) = Acos(ωt + ϕ) q k is the angular frequency and ϕ is the phase angle. where ω = m Step 2: Apply the initial condition x(0) = A. A = Acos(ϕ) Since cos(ϕ) = 1, we have ϕ = 0. Thus, the position function becomes: x(t) = Acos(ωt) Step 3: Find the maximum speed of the block. The speed of the block at any time t is given by: v(t) = −Aωsin(ωt) The maximum speed of the block is when sin(ωt) takes its maximum value of 1. Thus, the maximum speed vmax is: vmax = −Aω Substitute ω = q k m to get: r vmax = −A k m Question 15 Question A particle undergoes simple harmonic motion with an amplitude of 5 cm and a period of 1.5 seconds. If the displacement of the particle is 2.5 cm at time t = 0, determine: 1. The equation of motion for the particle. 2. The velocity of the particle at time t = 0.75 seconds. 3. The acceleration of the particle when it is 3 cm from the equilibrium position. 12 Solution 1. To find the equation of motion for the particle, we start with the general equation for simple harmonic motion: x(t) = A sin(ωt + ϕ) where: - A is the amplitude, - ω is the angular frequency, - ϕ is the phase angle. Given: Amplitude, A = 5 cm, Period, T = 1.5 s. We know that T = 2π ω , so we can solve for ω: 1.5 = 2π ω 2π 4π = 1.5 3 Now, we need to find the phase constant ϕ using the initial condition: At time t = 0, x(0) = 2.5 cm. Substituting this into the equation of motion: ω= x(0) = A sin(ϕ) = 2.5 5 sin(ϕ) = 2.5 2.5 = 0.5 5 π ϕ = sin−1 (0.5) = 6 Therefore, the equation of motion for the particle is: 4π π x(t) = 5 sin t+ 3 6 sin(ϕ) = 2. The velocity of the particle is given by the derivative of the displacement function: 4π π dx 4π =5 t+ v(t) = cos dt 3 3 6 At time t = 0.75 seconds: 4π 4π π v(0.75) = 5 cos × 0.75 + 3 3 6 4π π v(0.75) = 5 cos 2π + 3 6 4π 13π v(0.75) = 5 cos 3 6 √ ! 4π 3 v(0.75) = 5 − 3 2 13 √ v(0.75) = −10π 3 cm/s √ Therefore, the velocity of the particle at t = 0.75 seconds is −10π 3 cm/s. 3. The acceleration of the particle is given by the second derivative of the displacement function: 2 4π 4π π d2 x sin t+ a(t) = 2 = −5 dt 3 3 6 When the particle is 3 cm from the equilibrium position, we have: 4π π x(t) = 5 sin t+ =3 3 6 π 3 4π t+ = sin 3 6 5 To find the acceleration when x = 3 cm, we substitute x = 3 into the acceleration equation: 2 4π a(t) = −5 sin 3 Question 16 Question A particle undergoes simple harmonic motion with an amplitude of 3 cm and a period of 2 seconds. If the displacement of the particle is 2 cm at time t = 0, find the displacement at time t = 1 second. Solution Given that the particle undergoes simple harmonic motion, we can express its displacement x as a function of time t using the equation: x(t) = A cos(ωt + ϕ) where A is the amplitude, ω is the angular frequency, and ϕ is the phase angle. Step 1: Find the angular frequency ω using the period: Since T = 2π ω , we 2π have ω = 2π = = π rad/s. T 2 Step 2: Determine the phase angle ϕ: Given that the displacement of the particle is 2 cm at time t = 0, we have: x(0) = 3 cos ϕ = 2 Solving this equation for ϕ, we find ϕ = cos−1 23 . Step 3: Express the displacement x as a function of time t: Putting everything together, the displacement function becomes: 2 −1 x(t) = 3 cos πt + cos 3 14 Step 4: Find the particle’s displacement at time t = 1 second: Substitute t = 1 into the displacement function: 2 −1 x(1) = 3 cos π + cos 3 Since cos(π + θ) = − cos(θ), the displacement simplifies to: 2 2 −1 x(1) = −3 cos cos = −3 = −2 cm 3 3 Therefore, the displacement of the particle at time t = 1 second is −2 cm. Question 17 Question A particle undergoes simple harmonic motion with an amplitude of 5 cm and a period of 2 seconds. If the particle is located at 3 cm at time t = 0, find the displacement function s(t) for the particle. Solution Step 1: Identify the parameters of the simple harmonic motion. Given: Amplitude, A = 5 cm, Period, T = 2 seconds, Initial displacement, s(0) = 3 cm, Displacement function, s(t) = A sin 2π T t + ϕ , where ϕ is the phase angle. Step 2: Determine the angular frequency. The angular frequency, ω, can be 2π calculated using the formula ω = 2π T . ω = 2 = π radians/second. Step 3: Find the phase angle, ϕ. Given that s(0) = 3 cm, we substitute t = 0 in the displacement function to find ϕ. s(0) = 5sin(ϕ) = 3. This implies that sin(ϕ) = 53 . From this, we find that ϕ = sin−1 35 . Step 4: Write the displacement function s(t). Therefore, the displacement function is s(t) = 5 sin πt + sin−1 53 . Question 18 Question A particle undergoes simple harmonic motion with an amplitude of 5 cm and a period of 2 seconds. If the maximum velocity of the particle is 10 m/s, determine the equation of motion for the particle. Solution Step 1: Identify the given values and parameters of the simple harmonic motion. Given: Amplitude A = 5 cm = 0.05 m Period T = 2 s Maximum velocity vmax = 10 m/s 15 Step 2: Find the angular frequency ω. The angular frequency can be calculated using the formula: 2π ω= T 2π ω= = π rad/s 2 Step 3: Determine the equation of motion for the simple harmonic motion. The general equation for simple harmonic motion is given by: x(t) = A cos(ωt + ϕ) where x(t) is the displacement at time t, A is the amplitude, ω is the angular frequency, and ϕ is the phase angle. Step 4: Find the phase angle ϕ. Since the maximum velocity occurs when the displacement is zero, at this point the particle is at the equilibrium position and the cosine term is at its maximum value of 1. Therefore, ϕ is 0. Step 5: Substitute the values into the equation of motion. Substitute A = 0.05 m, ω = π rad/s, and ϕ = 0 into the equation of motion to get: x(t) = 0.05 cos(πt) Therefore, the equation of motion for the particle undergoing simple harmonic motion is x(t) = 0.05 cos(πt). Question 19 Question A particle of mass m is attached to a spring with spring constant k. Initially, the particle is at rest at the equilibrium position. At time t = 0, the particle is given an initial velocity v0 in the positive x-direction. Determine the maximum elongation of the spring during the subsequent motion. Solution Let’s denote the equilibrium position of the particle as x = 0. At any time t, the elongation of the spring from the equilibrium position is given by x(t). Step 1: Write the equation of motion for the particle. The equation of motion for simple harmonic motion is given by m d2 x = −kx dt2 Given that the particle is given an initial velocity v0 in the positive x-direction at t = 0, we have the initial conditions x(0) = 0 and 16 dx dt = v0 t=0 Step 2: Solve the differential equation. To solve this second-order differential equation, we can guess a solution of the form x(t) = A cos(ωt) + B sin(ωt) and find ω using k/m = ω 2 . Differentiating x(t) twice with respect to t and substiting back into the differential equation will give us the values of A and B. Step 3: Find the maximum elongation of the spring. The maximum elongation of the spring occurs at the amplitude of the oscillation. In this case, the amplitude is the maximum positive value of x(t), which can be found once A and B are determined. After solving for A and B, we can write down the equation for x(t) and then determine the maximum elongation from the amplitude of the resulting oscillatory function. Question 20 Question A mass m is attached to a spring with spring constant k. The mass is displaced from its equilibrium position and released. If the time period of the resulting simple harmonic motion is 5 seconds, find the frequency of the oscillation in Hz. Solution Step 1: Recall that the frequency (f ) of an oscillation is the reciprocal of the time period (T ), i.e., f = T1 . Step 2: Given that the time period T = 5 seconds, we can find the frequency f by substituting T into the formula: f = 15 Hz. Step 3: Therefore, the frequency of the oscillation of the mass attached to the spring is 15 Hz. Question 21 Question A particle undergoes simple harmonic motion with an amplitude of 0.2 m and a frequency of 2 Hz. If the displacement of the particle is given by x = 0.1 cos(4πt), where x is in meters and t is in seconds, determine the phase angle, period, angular frequency, and maximum velocity of the particle. Solution Step 1: To determine the phase angle, we compare the equation of motion with the general equation of simple harmonic motion. The general equation is given by x = A cos(ωt + ϕ), where A is the amplitude, ω is the angular frequency, and ϕ is the phase angle. By comparing the two equations, we see that ϕ = 0. 17 Step 2: The period of the motion can be found using the relation T = f1 , where T is the period and f is the frequency. Substituting the values given, we get T = 12 = 0.5 s. Step 3: The angular frequency is related to the frequency by the equation ω = 2πf . Substituting the frequency, we find ω = 2π × 2 = 4π rad/s. Step 4: The maximum velocity of the particle occurs when the displacement is zero. The velocity of a particle undergoing simple harmonic motion is given by v = −Aω sin(ωt + ϕ). At the equilibrium position (x = 0), the velocity is maximized. Substituting the values A = 0.1, ω = 4π, ϕ = 0, we find that the maximum velocity is 0.4π m/s. Question 22 Question A particle of mass 0.5 kg is attached to a spring with a spring constant of 20 N/m. It is initially displaced 0.1 m from its equilibrium position and released from rest. Find an expression for the position of the particle as a function of time. Solution Step 1: From Hooke’s Law, the force acting on the particle at any time t is given by F = −kx, where k is the spring constant and x is the displacement from equilibrium. Step 2: According to Newton’s Second Law, the acceleration of the particle F is given by a = , where m is the mass of the particle. m Step 3: Substituting F = −kx into the equation for acceleration, we have k a = − x. m Step 4: The acceleration can also be expressed as the second derivative of d2 x displacement with respect to time, i.e., a = 2 . dt k d2 x Step 5: Equating the two expressions for acceleration, we get 2 = − x. dt m Step 6: This is a second-order linear differential equation with constant coefficients, the general solution of which is x(t) =rA cos(ωt) + B sin(ωt), where k A and B are constants to be determined and ω = is the angular frequency. m Step 7: Applying the initial conditions x(0) = 0.1 and v(0) = 0, where dx v(t) = is the velocity of the particle, we can find the values of A and B. dt Step 8: After finding A and B, the final expression for the position of the particle as a function of time is x(t) = A cos(ωt) + B sin(ωt). 18 Question 23 Question A particle undergoes simple harmonic motion with an amplitude of 3 cm and a frequency of 2 Hz. If the particle is at a displacement of 2 cm at time t = 0, determine the equation for the displacement of the particle as a function of time. Solution Given parameters: Amplitude, A = 3 cm Frequency, f = 2 Hz Displacement at t = 0, s(0) = 2 cm The general form of the displacement equation for simple harmonic motion is: s(t) = A sin(ωt + ϕ) where ω = 2πf is the angular frequency and ϕ is the phase angle. Step 1: Find the angular frequency, ω ω = 2πf = 2π × 2 = 4π rad/s Step 2: Find the phase angle, ϕ Given s(0) = 2 cm: s(0) = A sin(ϕ) = 3 × sin(ϕ) = 2 Solving for ϕ: 2 3 2 −1 ⇒ ϕ = sin 3 sin(ϕ) = Step 3: Write the equation of motion The equation for the displacement of the particle as a function of time is: s(t) = 3 sin 4πt + sin−1 (2/3) Question 24 Question A mass-spring system is oscillating with an amplitude of 0.2 m. If the maximum velocity of the mass is 0.5 m/s, determine the frequency of the oscillation. 19 Solution Step 1: We know that the maximum velocity of a mass-spring system in simple harmonic motion is given by vmax = Aω, where A is the amplitude and ω is the angular frequency. Step 2: Given that vmax = 0.5 m/s and A = 0.2 m, we can write 0.5 = 0.2ω. Step 3: Solving for ω, we find ω = 0.5 0.2 = 2.5 rad/s. Step 4: The frequency f of simple harmonic motion is related to the angular frequency ω ω by the formula f = 2π . Step 5: Substituting ω = 2.5 rad/s into the formula 2.5 gives f = 2π ≈ 0.3979 Hz. Step 6: Therefore, the frequency of the oscillation is approximately 0.3979 Hz. Question 25 Question A block of mass m is attached to a spring with spring constant k. The block is pulled to a distance A from the equilibrium position and released. Determine the maximum speed of the block during the oscillation. Solution Let’s denote the equilibrium position of the block as x = 0, the initial position as x = A, and the maximum displacement from equilibrium as x = ±A. The potential energy of the block-spring system at any position x is given by P E = 1 2 2 kx . Step 1: The total mechanical energy E of the system is conserved and is the sum of kinetic and potential energies. E = KE + P E Initially, when the block is at position x = A, all energy is potential energy, hence E = P E(A) = 21 kA2 . Step 2: When the block reaches the equilibrium position at x = 0, all 2 energy is kinetic energy. Therefore, E = KEmax = 21 mvmax . Step 3: Since energy is conserved, the total energy at x = A must equal the total energy at x = 0. 1 2 1 2 kA = mvmax 2 2 Step 4: Solving for the maximum speed vmax of the block: r k 2 vmax = A m q k 2 A . Hence, the maximum speed of the block during oscillation is vmax = m 20 Question 26 Question A mass-spring system has a mass of 0.5 kg attached to a spring with a spring constant of 20 N/m. If the maximum velocity of the mass is 1 m/s, find the amplitude of the oscillation. Solution Step 1: Find the angular frequency of the oscillation using the formula: r k ω= m where ω is the angular frequency, k is the spring constant, and m is the mass. r √ √ 20 ω= = 40 = 2 10 ≈ 6.32 rad/s 0.5 Step 2: The maximum velocity in simple harmonic motion is given by the formula: vmax = ωA where vmax is the maximum velocity and A is the amplitude of the oscillation. Given vmax = 1 m/s, we can rearrange the formula to solve for A: A= vmax 1 1√ = √ = 10 ≈ 1.58 m ω 2 2 10 Therefore, the amplitude of the oscillation is approximately 1.58 m. Question 27 Question A mass m is attached to a spring with spring constant k. The mass is released from rest at its equilibrium position. It reaches its maximum displacement of A from the equilibrium position, then returns to its equilibrium position before continuing to the maximum displacement of A in the opposite direction. Determine the period of oscillation of the mass. Solution To determine the period of oscillation, we first need to find the angular frequency ω, and then we can use the formula for the period T = 2π ω . Step 1: Find the angular frequency ω. The angular frequency ω can be found using the formula: r k ω= m 21 Step 2: Find the period T . The period of oscillation, T , is given by: T = Since ω = q k m, 2π ω we have: 2π T = q = 2π k m r m k Therefore, the period of oscillation of the mass is 2π pm k . Question 28 Question A particle of mass 0.5 kg is attached to a spring with a spring constant of 20 N/m. Initially, the particle is displaced 0.1 m from its equilibrium position and released from rest. Determine the amplitude, frequency, angular frequency, and period of the resulting simple harmonic motion. Solution Step 1: Find the amplitude A A = 0.1 m Step 2: Find the angular frequency ω r r √ k 20 ω= = = 40 rad/s m 0.5 Step 3: Find the frequency f √ ω 40 6.32 f= = ≈ Hz 2π 2π 2π Step 4: Find the period T T = 1 1 1 = ≈ s ≈ 0.995 s f 6.32/2π 1.005 Therefore,√the amplitude of the simple harmonic motion is 0.1 m, the angular frequency is 40 rad/s, the frequency is approximately 6.32 2π Hz, and the period is approximately 0.995 s. 22 Question 29 Question A particle undergoes simple harmonic motion with an amplitude of 5 cm and a frequency of 2 Hz. If the particle is at its equilibrium position at time t = 0, determine the displacement, velocity, and acceleration of the particle at time t = 32 seconds. Solution Step 1: Find the angular frequency ω The angular frequency ω is related to the frequency f by the equation ω = 2πf . Given that f = 2 Hz, we can compute ω by ω = 2π × 2 = 4π rad/s Step 2: Determine the displacement at t = 23 s The displacement at time t for a particle undergoing simple harmonic motion is given by x(t) = A sin(ωt + ϕ) where A is the amplitude, ω is the angular frequency, t is time, and ϕ is the phase angle. Since the particle is at its equilibrium position at t = 0, we have ϕ = 0. Thus, the displacement at time t = 32 s is 2 2 8π x = 5 sin 4π × = 5 sin 3 3 3 √ √ 2π 3 5 3 2 = 5 sin =5× = cm x 3 3 2 2 Step 3: Calculate the velocity at t = The velocity of the particle is given by 2 3 s v(t) = Aω cos(ωt + ϕ) Substitute A = 5, ω = 4π, t = 23 , and ϕ = 0: 2 2 8π v = 5 × 4π cos 4π × = 20π cos 3 3 3 2 2π 1 v = 20π cos = −20π × = −10π cm/s 3 3 2 Step 4: Find the acceleration at t = 32 s The acceleration of the particle is given by a(t) = −Aω 2 sin(ωt + ϕ) 23 Substitute A = 5, ω = 4π, t = 23 , and ϕ = 0: 2 2 8π a = −5 × (4π)2 sin 4π × = −80π 2 sin 3 3 3 √ √ 3 2 2π 2 a = −80π 2 sin = −80π 2 × = −40π 2 3 cm/s 3 3 2 Therefore, at t = 23 s, the displacement is √ and acceleration is −40π 2 3 cm/s2 . √ 5 3 2 cm, velocity is −10π cm/s, Question 30 Question An object of mass m is attached to a spring with spring constant k. The object is displaced from its equilibrium position by a distance x0 and released from rest. Find the maximum speed of the object during its motion. Solution Step 1: Determine the maximum potential energy of the object. The maximum potential energy occurs when the object is at its maximum displacement x0 . The potential energy at this position is given by P E = 21 kx20 . Step 2: Use conservation of mechanical energy to find the maximum speed. At the equilibrium position, all the potential energy is converted to kinetic energy. Therefore, the maximum potential energy at x0 equals the maximum kinetic energy at the equilibrium position. The kinetic energy is given by KE = 1 2 2 mvmax , where vmax is the maximum speed of the object. Step 3: Equate the potential energy and kinetic energy expressions. 2 We have 12 kx20 = 12 mvmax . Step 4: Solve for the maximum speed vmax . q q x0 k 2 k From the equation in Step 3, we find vmax = m x0 = x0 m . Therefore, the maximum speed of the object during its motion is vmax = q k m. Question 31 Question A mass attached to a spring executes simple harmonic motion with an amplitude of 5 cm and a period of 2 seconds. If the maximum acceleration of the mass is 2 8 m/s , determine the mass and the spring constant. 24 Solution Step 1: Identify the formula for the period of simple harmonic motion: The period (T ) of simple harmonic motion is related to the angular frequency (ω) by the equation: 2π T = ω Step 2: Determine the angular frequency from the period: Given that the period (T ) is 2 seconds, we can find the angular frequency (ω) using the equation from Step 1: 2π 2π = = π rad/s ω= T 2 Step 3: Determine the maximum velocity from the amplitude and period: The maximum velocity in simple harmonic motion is given by: vmax = ω · amplitude Given that the amplitude is 5 cm (or 0.05 m), we can find the maximum velocity: vmax = π × 0.05 = 0.157 m/s Step 4: Determine the mass using the maximum acceleration: The acceleration in simple harmonic motion is related to the angular frequency and the amplitude by: amax = ω 2 · amplitude 2 Given that the maximum acceleration is 8 m/s , we can find the mass (m): m= amax 8 = 2 ≈ 0.81 kg ω2 π Step 5: Determine the spring constant using the mass and maximum acceleration: The spring constant (k) can be found using the equation: k= m · amax 0.81 × 8 = = 129.6 N/m amplitude 0.05 Therefore, the mass of the object is approximately 0.81 kg and the spring constant is approximately 129.6 N/m. Question 32 Question A 0.5 kg mass attached to a spring oscillates with a frequency of 2 Hz. If the amplitude of the oscillation is 0.1 m, what is the maximum speed of the mass during the oscillation? 25 Solution Step 1: We know that the frequency of oscillation, f , is related to the angular ω frequency, ω, by the formula f = 2π . Therefore, we can find ω using the given frequency. Step 1: f = 2 Hz ω = 2πf = 2π × 2 = 4π rad/s Step 2: The maximum speed of the mass during simple harmonic motion is given by the formula vmax = ωA, where A is the amplitude of oscillation. Step 2: A = 0.1 m vmax = 4π × 0.1 = 0.4π m/s Therefore, the maximum speed of the mass during the oscillation is 0.4π ≈ 1.26 m/s. Question 33 Question A spring-mass system is set into oscillation with an amplitude of 10 cm and a frequency of 2 Hz. If the maximum speed of the mass is 20 cm/s, determine the displacement of the mass when its speed is 10 cm/s. Solution Step 1: Determine the angular frequency (ω) using the formula f = ω 2π . ω = 2π × f = 2π × 2 = 4π rad/s Step 2: Calculate the maximum displacement of the mass using the formula A = Vmax ω , where A is the amplitude and Vmax is the maximum speed. 20 5 ⇒ 10 = rad 4π 2π Step 3: Determine the displacement when the speed is 10 cm/s. Let the displacement be x. 5t 5t ⇒ cos =1 10 = 10 cos(ωt) = 10 cos 2π 2π 10 = Step 4: Find the corresponding time t for speed 10 cm/s. 5t =0⇒t=0s 2π Step 5: Substitute the time into the equation for displacement. x = 10 cos(0) = 10 cm Therefore, the displacement of the mass when its speed is 10 cm/s is 10 cm. 26 Question 34 Question A particle undergoes simple harmonic motion with an amplitude of 5 cm and a period of 2 seconds. If the displacement of the particle is 3 cm at time t = 1 second, find the equation of motion for the particle. Solution Step 1: Determine the angular frequency ω 2π We know that the angular frequency ω = , where T is the period. Given T that T = 2 seconds, we have ω= 2π = π rad/s 2 Step 2: Determine the equation of motion The general equation of motion for simple harmonic motion is given by x(t) = A cos(ωt + ϕ), where: - A is the amplitude of the motion, - ω is the angular frequency, - t is time, and - ϕ is the phase angle. Given that the amplitude A = 5 cm and the particle is at 3 cm at t = 1 second, we can substitute these values into the general equation and solve for the phase angle ϕ: 3 = 5 cos(π + ϕ) Step 3: Solve for ϕ First, we need to find the quadrant in which the particle is located at t = 1 second. Since the particle is at 3 cm and the amplitude is 5 cm, the particle is at the negative end of its motion. Therefore, cos(π) = −1, and we have 3 = 5(−1) = −5 Solving for ϕ: −5 = 5 cos(ϕ) cos(ϕ) = −1 ϕ=π Step 4: Write the equation of motion Now we have found the phase angle ϕ = π. Substituting this into the general equation, we get: x(t) = 5 cos(πt + π) Therefore, the equation of motion for the particle is x(t) = 5 cos(πt + π). 27 Question 35 Question A particle is undergoing simple harmonic motion with an amplitude of 6 cm and a frequency of 2 Hz. If at t = 0 s the particle is at its maximum displacement from the equilibrium position of 3 cm in the positive direction, determine the equation of motion for the particle. Solution Step 1: Determine the angular frequency ω. Given that the frequency is 2 Hz, ω to find ω. we can use the relation f = 2π f = 2 Hz = ω 2π ω = 4π rad/s Step 2: Write the equation of motion for simple harmonic motion. The equation of motion for simple harmonic motion is given by: x(t) = A cos(ωt + ϕ) where: - x(t) is the displacement of the particle at time t, - A is the amplitude of the motion, - ω is the angular frequency, - ϕ is the phase angle. Step 3: Substitute the given values and determine the phase angle, ϕ. With the given information, we have: - Amplitude, A = 6 cm - Initial displacement, x(0) = 3 cm Plugging in these values to the general equation of motion: x(0) = A cos(ϕ) = 3 6 cos(ϕ) = 3 cos(ϕ) = 1 2 π rad 3 Step 4: Write the equation of motion for the particle. Finally, plugging in the known values into the equation of motion: π x(t) = 6 cos 4πt + 3 ϕ= 28