Question

For a particle performing SHM:

• A. The kinetic energy is never equal to the potential energy
• B. The kinetic energy is always equal to the potential energy
• C. The average kinetic energy in one time period is equal to the average potential energy in this period.
• D. The average kinetic energy in any time interval is never equal to average potential energy in that interval.

Answer 1

Answer: (C)

The average kinetic energy in one time period is equal to the average potential energy in this period.

For SHM,

$x=A\cos (\omega t+\varphi )v=-A\omega \sin (\omega t+\varphi )$

Kinetic energy$=\frac{1}{2}m{v}^2$

$=\frac{1}{2}m{A}^2{\omega }^2{\sin }^2(\omega t+\varphi )$

Potential energy $=\frac{1}{2}k{x}^2$

$=\frac{1}{2}k{A}^2{\cos }^2(\omega t+\varphi )$

$=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2(\omega t+\varphi )[K=m{\omega }^2]$

$⟨KE⟩=\frac{1}{2}m{A}^2{\omega }^2⟨{\sin }^2(\omega t+\varphi )⟩⟨PE⟩=\frac{1}{2}m{A}^2{\omega }^2⟨{\cos }^2(\omega t+\varphi )⟩⟨{\sin }^2(\omega t+\varphi )⟩=\frac{1}{T}{\int }_0^T{\sin }^2(\omega t+\varphi )dt=\frac{\omega }{2\pi }{\int }_0^{2\pi /\omega }{\sin }^2(\omega t+\varphi )dt=\frac{1}{2}⟨{\cos }^2(\omega t+\varphi )⟩=\frac{1}{2}⟨KE⟩=⟨PE⟩=\frac{1}{4}m{A}^2{\omega }^2$