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

2024
Views:

97

Pages:

5
Author:

Jabin F.

Key Points: At equilibrium position, speed is maximum, and acceleration is minimum. At extreme position, acceleration is maximum, and speed is minimum. In SHM, time period of oscillation is independent of energy or amplitude. The time taken to go from mean to extreme position or extreme to mean position is T/4, and from extreme to extreme is T/2. Formulas: Force (F) = -kx Potential energy (U) = (1/2)kx^2 + c Time period (T) ∝ √(m/k) Position (x) = A sin(ωt + φ) Velocity (v) = ω√(A^2 - x^2) Example: When a particle is released in a smooth tunnel made along a chord of Earth, it executes SHM. If T is the time period of oscillation, then a particle released at point 1, 2, or 3 will reach the center (0) after time T/4. Problem 1: Equation of motion: mẍ + δẋ - mg = 0 Initial velocity: v₀ Terminal velocity: vₜ Solution: Set acceleration (ẍ) to zero for terminal velocity. Solve for vₜ: vₜ = mg/δ Problem 2: Find the maximum force F such that the block does not move. Solution: Apply vertical equilibrium: N = mg + Fsin(60°) Apply horizontal equilibrium: Fcos(60°) = μN Substitute N from the first equation into the second equation. Solve for F: F = 30 N Problem: A particle is at rest at x = 0. A force F = (1 - 9x)i is applied on the particle. Find the maximum speed attained by the particle. Solution: Equation of motion: 1 - 9x = ma Integrate both sides: ∫(1 - 9x)dx = m∫dv Solve the integrals: x - (9/2)x^2 = (1/2)mv^2 Rearrange the equation: v^2 = (1/9)(x - x^2) For v to be maximum, v^2 must also be maximum. Differentiate v^2 with respect to x and set it to zero: dv^2/dx = (1/9)(1 - 2x) = 0 Solve for x: x = 1/2 Substitute x = 1/2 into the equation for v^2: v^2 = (1/9)(1/2 - (1/2)^2) = 1/36 Take the square root of both sides: v = 1/6