Driving Function, Forcing Function, or Input

Definition: S uppose we h ave t he f ollowing s econd-order d ifferential eq uation wi th co nstant coefficients a long w ith i nitial c onditions s pecified at a ti me th at we sh all ta ke to be t = 0: 2 ′ a ddt2y + b dy dt + cy = g(t), y(0) = y0 , y (0) = y1 . The function g is called the driving function, forcing function, or input of the system. A solution y(t) of the differential equation on an interval I containing t = 0 that satisfies the initial conditions is called the response or output of the system. Hooke’s Law: Suppose that a flexible spring is suspended vertically from a rigid support and then a mass m is attached to its free end. The spring exerts a restoring force F opposite to the direction of elongation and proportional to the amount of elongation s. So, F = −ks, where k > 0. Definition: k is called the spring constant. Newton’s Second Law and Free Motion: If the mass vibrates free of all other external forces - free motion - then Newton’s Second Law gives 2 m ddt2x = −kx. Differential Equation of Free Undamped Motion: The second-order differential equation d2 x dt2 + k mx =0 describes simple harmonic motion of free undamped motion. Alternative Form: The above equation can be re-written as d2 x dt2 + ω 2 x = 0, where ω 2 = k/m. Solution: The general solution of the differential equation is x(t) = c1 cos ωt + c2 sin ωt. Initial Conditions: The initial conditions associated with the equation are x(0) = x0 and x′ (0) = x1 , the initial displacement and initial velocity of the mass, respectively. For example, if x0 > 0, x1 < 0, the mass starts from a point below the equilibrium position with an imparted upward velocity. If x1 > 0, the mass starts with an imparted downward velocity. Alternative Form of x(t): The alternative form of the solution is x(t) = A sin(ωt + ϕ), where A = p c21 + c22 and ϕ is a phase angle defined by tan ϕ = c1 c2 . Definition: The period of motion is T = 2π/ω. The number T represents the time (measured in seconds) it takes the mass to execute one cycle of motion. Definition: The extreme displacement of the mass is the positive displacement corresponding to the mass attaining its greatest distance below the equilibrium position or the negative displacement corresponding to the mass attaining its greatest height above the equilibrium position. Definition: The frequency of motion is f = 1/T = ω/2π and is the number of cycles completed each second. Definition: The number ω is called the circular frequency of the system. Definition: ω/2π is referred to as the natural frequency of the system. Exercise 1. A 1/8-kg mass is attached to a spring with stiffness k = 16 N/m. The mass is displaced 1/2 m to the right of the equilibrium point and given an outward velocity (to the right) √ of 2 m/sec. Neglecting any damping or external forces that may be present, determine the equation of motion of the mass along with its amplitude, period, and natural frequency. How long after release does the mass pass through the equilibrium position? (Nagle Section 4.9 Example 1) Exercise 2. A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from the equilibrium position with a downward velocity of 2 ft/s from the equilibrium position. Find the equation of motion. (#4) Class Exercise 1. A force of 400 newtons stretches a spring 2 meters. A mass of 50 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 10 m/s. Find the equation of motion. (#6) Class Exercise 2. A 20-kilogram mass is attached to a spring. If the frequency of simple harmonic motion is 2/π cycles/s, what is the spring constant k? What is the frequency of simple harmonic motion if the original mass is replaced with an 80-kilogram mass? (#2) Exercise 3. A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 2 ft/s. How many complete cycles will the mass have completed at the end of 4π seconds? (#8) Class Exercise 3. A mass weighing 10 pounds stretches a spring 14 foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point 13 foot above the equilibrium position with a downward velocity of 54 ft/s. Express the equation of motion in the form x(t) = A sin(ωt + ϕ). (#10(a)) Class Exercise 4. A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is√initially released from a point 1 foot above the equilibrium position with an upward velocity of 3 ft/s. Find the times at which the mass is heading downward at a velocity of 3 ft/s. (#12) Differential Equation of Free Damped Motion: The differential equation of free damped motion is 2 m ddt2x = −kx − β dx dt , where β is a damping constant. Alternative Form: The above equation can be re-written as d2 x dt2 where 2λ = β m and ω 2 = 2 + 2λ dx dt + ω x = 0, k m. Case 1: λ2 − ω 2 > 0: The system is overdamped and the general solution is x(t) = e−λt (c1 e √ λ2 −ω 2 t √ λ2 −ω 2 t + c2 e− ). Case 2: λ2 − ω 2 = 0: The system is critically damped and the general solution is x(t) = e−λt (c1 + c2 t). Case 3: λ2 − ω 2 < 0: The system is underdamped and the general solution is √ √ x(t) = e−λt (c1 cos ω 2 − λ2 t + c2 sin ω 2 − λ2 t). The above equation can be rewritten as √ x(t) = Ae−λt sin( ω 2 − λ2 t + ϕ), where A = p c21 + c22 and ϕ is a phase angle defined by tan ϕ = c1 c2 . Definition: The coefficient Ae−λt is called the damped amplitude. √ Definition: The number 2π/ ω 2 − λ2 is called the quasi period. The quasi period is the time interval between two successive maxima of x(t). Definition: The number √ ω 2 − λ2 /2π is the quasi frequency. Exercise 4. Assume that the motion of a mass-spring system with damping is governed by d2 y dt2 ′ + b dy dt + 25y = 0; y(0) = 1, y (0) = 0. Find the equation of motion and sketch its graph for the three cases when b = 6, 10, and 12. (Nagle Section 4.9 Example 2) Exercise 5. A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached √ to it. The medium through which the mass moves offers a damping force numerically equal to 2 times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 5 ft/s. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant? (#26) Class Exercise 5. After a mass weighing 10 pounds is attached to a 5-foot spring, the spring measures 7 feet. The mass is removed and replaced with another mass that weighs 8 pounds. The entire system is placed in a medium that offers a damping force that is numerically equal to the instantaneous velocity. (#30) (a) Find the equation of motion if the mass is initially released from a point 12 foot below the equilibrium position with a downward velocity of 1 ft/s. √ (b) Express the equation of motion in the form x(t) = Ae−λt sin ( ω 2 − λ2 t + ϕ). (c) Find the times at which the mass passes through the equilibrium position heading downward. (d) Graph the equation of motion. Class Exercise 6. A mass weighing 24 pounds stretches a spring 4 feet. The subsequent motion takes place in a medium that offers a damping force numerically equal to β (β > 0) times the instantaneous velocity. If the mass is initially √ released from the equilibrium position with an upward velocity of 2 ft/s, show that when β > 3 2 the equation of motion is p 2 −2βt/3 e sinh β 2 − 18t. x(t) = √ −3 3 2 β −18 (#32) Differential Equation of Driven Motion with Damping: Here is the differential equation of driven or forced motion: 2 m ddt2x = −kx − β dx dt + f (t), where f (t) is an external force acting on the mass. Alternative Form: The above equation can be rewritten as: d2 x dt2 2 + 2λ dx dt + ω x = F (t), where F (t) = f (t)/m, 2λ = β/m, and ω 2 = k/m. Exercise 6. A 10-kg mass is attached to a spring with stiffness k = 49 N/m. At time t = 0, an external force f (t) = 20 cos 4t N is applied to the system. The damping constant for the system is 3 N-sec/m. Determine the steady-state solution for the system. (Nagle Section 4.10 Example 1) Exercise 7. A mass of 1 slug is attached to a spring whose constant is 5 lb/ft. Initially, the mass is released 1 foot below the equilibrium position with a downward velocity of 5 ft/s, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 2 times the instantaneous velocity. (#34) (a) Find the equation of motion if the mass is driven by an external force equal to f (t) = 12 cos 2t + 3 sin 2t. (b) Graph the transient and steady-state solutions on the same coordinate axes. (c) Graph the equation of motion. Class Exercise 7. A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at t = 0, an external force equal to f (t) = e−t sin 4t is applied to the system. Find the equation of motion if the surrounding medium offers a damping force that is numerically equal to 8 times the instantaneous velocity. Analyze the displacements for t → ∞. (#36) Class Exercise 8. When a mass of 2 kilograms is attached to a spring whose constant is 32 N/m, it comes to rest in the equilibrium position. Starting at t = 0, a force equal to f (t) = 68e−2t cos 4t is applied to the system. It can show that the equation of motion is x = − 21 cos 4t + 9 4 sin 4t + 12 e−2t cos 4t − 2e−2t sin 4t. Find the equation of the motion in the form x(t) = A sin(ωt + ϕ) + Be−2t sin(4t + θ). What is the amplitude of vibrations after a very long time? (#38)