University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

122

Pages:

6
Author:

Jabin F.

Probability of Finding a SHM: The probability of finding a Simple Harmonic Oscillator (SHO) between positions x and x+dx is given by: P = (1/T) * dt where T is the time period. The time period T is related to the angular frequency ω and the displacement x: T = 2π/ω = 2π * sqrt(m/k) The probability can be expressed in terms of dx: P = (1/T) * (dx/v) = (1/(2π)) * (sqrt(m/k)) * (dx/x) Bottom Section: Q-5: Problem Statement: A particle is dropped from a height H. In the absence of air resistance, it reaches the ground with speed v₀. If a drag force of magnitude f = kv² acts on it, find the final speed near the ground. Solution: Without drag force: The final speed is given by v₀² = 2gH. With drag force: The equation of motion (EOM) is: mg - kv² = ma Rearranging and integrating the equation, we get: ∫(v₀ to v) dv / (1 - (v²/v₀²)) = ∫(0 to H) g dx Solving this integral and simplifying, we can find the final speed v near the ground. Top Section (Continuation of Q-5): The solution to the problem of a particle falling under the influence of gravity and a drag force continues with the following steps: -9gH/v₀² = ln(1 - v²/v₀²) This equation is simplified and solved for the final velocity v: v = v₀ * sqrt((e - 1) / e) Bottom Section (Q-6): Problem Statement: A block is pulled with a constant force F until it reaches maximum displacement, at which point the force is removed. The task is to find the maximum speed attained by the block during its return motion. Solution: Equation of Motion (EOM): F - kx = ma This equation relates the applied force F, the spring force kx, the mass m, and the acceleration a of the block. Work-Energy Principle: The work done by the force F is equal to the change in kinetic energy of the block. ∫(F - kx) dx = (1/2)mv² Integrating and solving for the maximum extension x_max, we get: x_max = 2F/k Conservation of Energy: At the maximum extension, the total mechanical energy (kinetic + potential) is conserved. (1/2)kx_max² = (1/2)mv² + 0 Substituting x_max and solving for v, we get the maximum speed: v = (2F/√(km)) Equation of Motion (EOM): The equation of motion for the particle is given by: mg - kv = ma Terminal Velocity: At terminal velocity, the acceleration is zero. Therefore, mg - kv_t = 0 Solving for the terminal velocity v_t, we get: v_t = mg/k Modified EOM: Substituting the expression for terminal velocity into the original EOM and rearranging, we get: dv/(mg - kv) = dt/m Integration: Integrating both sides of the equation, we obtain: -ln(mg - kv) / k = t/m + C Applying the initial condition v = 0 at t = 0, we find the integration constant C. Solving for t₁ and t₂: Using the given conditions for velocities at times t₁ and t₂, we can set up two equations and solve for t₁ and t₂. Finding the Ratio t₂/t₁: Finally, the ratio t₂/t₁ is calculated, and the result is simplified to obtain the final answer.