Question

The potential energy of a particle executing SHM changes from maximum to minimum in $5\text{s}$. The time period of SHM is:

• A. $17\text{s}$
• B. $18\text{s}$
• C. $19\text{s}$
• D. $20\text{s}$

Answer 1

The correct option is D $20\text{s}$

Given, the time of oscillation of potential energy between maximum and minimum is $5\text{s}$. So, the time of oscillation of potential energy between two successive maxima will be $10\text{s}$.
Hence, frequency,
$f=\frac{1}{10}\text{Hz}$
We know that potential energy oscillates with double the frequency of the SHM.
So, the frequency of SHM,
$f_o=\frac{1}{2}\times \frac{1}{10}=\frac{1}{20}\text{Hz}$
Now, time period of SHM,
$T_o=\frac{1}{f_o}=20\text{s}$

Alternative:
PE changes from maximum to minimum in $T/4$, where $T$ is time period of SHM.
Since $\frac{T}{4}=5$
$\Rightarrow T=20\text{s}$