JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Elasticity & SHM

A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about

x = 0

. When its potential energy (PE) equals kinetic energy (KE), the position of the particle will be

Answer: (C)

Solution

In SHM the total energy is fixed at

\frac{1}{2} m ω^{2} A^{2}

, split as

P E = \frac{1}{2} m ω^{2} x^{2}

and

K E = \frac{1}{2} m ω^{2} (A^{2} - x^{2})

. Demanding

P E = K E

gives

x^{2} = A^{2} - x^{2}

, so

2 x^{2} = A^{2}

and

x = \frac{A}{2}

. At that displacement the energy is shared equally between potential and kinetic forms.

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option