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. $T_{1}=2 T_{2}$
• C. $2 T_{1}=T_{2}$
• D. None of these

Answer 1

Answer: (C)

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