Question

The potential energy of a particle ($U_{x}$) executing S.H.M. is given by
(a) $U_{x}=\frac{k}{2}(x-a)^{2}$
(b) $U_{x}=k_{1}x+k_{2}x^{2}+k_{3}x^{3}$
(c) $U_{x}=Ae^{-bx}$
(d) $U_{x}=a\ constant$

Answer 1

The potential energy (P.E.) of a particle performing S.H.M. potential energy is the energy possessed by the particle when it is at rest. Thus it is calculated as -

Considering a particle of mass m performing simple harmonic motion at a distance x from its mean position. The restoring force acting on the particle is F= -kx where k is the force constant.

dw = – fdx = – (- kx)dx = kxdx

Therefore, the total work done to displace the particle now from 0 to x is

∫dw=  ∫kxdx = k ∫x dx

Hence Total work done = 1/2Kx² = 1/2 m ω²x²

Therefore Potential energy = 1/2kx² = 1/2 m ω²x²

Whereas for kinetic -

Considering a particle with mass m performing simple harmonic motion along with a path.

v= ±ω √a² – x²

v² = ω² ( a² – x²)

Kinetic energy= 1/2 mv2  = 1/2 mω2 ( a²  – x²)

As, k/m = ω²

Therefore k = m ω²

Kinetic energy= 1/2 k( a² – x²)