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

2024
Views:

280

Pages:

5
Author:

Jabin F.

# Unstable equilibrium point (USEP): The equilibrium point to which particle doesn't return on being slightly displaced is called USEP. at USEP, potential energy is max. i.e., \[\frac{\partial^{2}V}{\partial x^{2}}|_{x=x_{0}} < 0 (-ve')\] # Time Energy Relation: When potential energy is given then energy remains conserved. E = KE + PE = ½mv² + V(x) = constant => E = ½m(dx/dt)² + V(x) => dx/dt = √(2/m[E - V(x)]) on integrating dt = √(m/2) * ∫ dx / √(E - V(x)) # Time Amplitude Relation -> E α A valid only for SHM & em waves. General case: E = ½mv² + V(x) E = ½mω² + kx^n at amplitude (x = A), v = 0 therefore E = kA^n ∴ t α A^(1-n/2) = (k^n)^(1-n/2) * A^(1-n/2) => t α A^(1-n/2) P-97 Q-35 A particle of mass m moves in 1d in the potential V(x) = kx^n, (k>0). At t = 0, the particle starts from the rest at x = A. For bounded motion, the time period of its motion is Solution: V(x) = kx^n -> n = 4 t α A^(1-n/2) = A^(1-4/2) = A^(-1) => t α A^(-1) Q-95 The energy of a one-dimensional system, governed by the Lagrangian L = ½mx² - ½kx^n, where k and n are two positive constants, is E. The time period of oscillation T satisfies: Solution: As L = KE - PE, so V(x) = ½kx^n => n = 2 ∴ T α (1/E)^(1/2) => T α E^(-1/2)