Question

For a body executing S.H.M.
$\left(a\right)$ Potential energy is always equal to its K.E
$\left(b\right)$ Average potential and kinetic energy over any given time interval are always equal .
$\left(c\right)$ Sum of the kinetic and potential energy at any point of time is constant.
$\left(d\right)$ Average K.E in one time period is equal to average potential energy on one time period.

Choose the most appropriate option from the options given below

• A. $\left(b\right)$ and $\left(c\right)$
• B. $\left(c\right)$ and $\left(d\right)$
• C. only $\left(b\right)$
• D. only $\left(c\right)$

Answer 1

The correct option is B $\left(c\right)$ and $\left(d\right)$

For a particle executing SHM,

Potential energy is given by $U=\frac{1}{2}m{\omega }^2{x}^2$
Kinetic energy is given by $K=\frac{1}{2}m{\omega }^2(A^2-x^2)$
Total energy is $U+K=K=\frac{1}{2}m{\omega }^2{A}^2$


As $U\ne K$ always
$\left(a\right)$ is incorrect

As Total energy is constant, $\left(c\right)$ is correct

Average Kinetic and potential energy over interval of one time period is $<K>=<U>=\frac{K{A}^2}{4}$
$\left(d\right)$ is correct and $\left(b\right)$ is incorrect