Question

In simple harmonic motion, let the time period of variation of potential energy is $T_1$ and time period of variation of position is $T_2$, then relation between $T_1$ and $T_2$ is

• A. $T_1=T_2$
• B. $T_1=2T_2$
• C. $2T_1=T_2$
• D. None of these

Answer 1

Answer: (C)

For SHM,
Displacement, $x(t)=A\cos (\omega t+\varphi )$
$\Rightarrow T_2=2\pi /\omega$
Potential energy, PE $=\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2{\cos }^2(\omega t+\varphi )\dots$ (ii)
$\Rightarrow =\frac{1}{2}k{A}^2\frac{[1+\cos 2(\omega t+\varphi )]}{2}$
$\therefore T_1=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$
$T_1=\frac{T_2}{2}$
$\Rightarrow 2T_1=T_2$
So, the relation between $T_1$ and $T_2$ is $2T_1=T_2$.