Question

A spring block system is performing $SHM$ with a time period of $4\text{s}$. If at any instant, the kinetic energy of the system is represented as $K_E$ and potential energy as $K_P$, then what is the time period of oscillation of $K_E-K_P$?

• A. $8\text{s}$
• B. $1\text{s}$
• C. $2\text{s}$
• D. $4\text{s}$

Answer 1

The correct option is C $2\text{s}$

Given that,
Time period of oscillation $T=4\text{s}$
We know that the frequency of oscillations of kinetic energy and potential energy of a $SHM$ is twice the frequency of $SHM$.
Hence,
the difference of kinetic energy and potential energy will also have the frequency of oscillations twice the frequency of $SHM$.
Given, the time period of $SHM$,
$T=4\text{s}$
So, the frequency of $SHM$,
$f=\frac{1}{4}{\text{s}}^{-1}$
Now, the frequency of difference of kinetic energy and potential energy,
$f^{{}^{\prime }}=2\times f=2\times \frac{1}{4}=\frac{1}{2}{\text{s}}^{-1}$
So, its time period,
$T^{{}^{\prime }}=\frac{1}{f^{{}^{\prime }}}=\frac{1}{1/2}=2\text{s}$.