Question

A particle is executing linear SHM. The average kinetic energy and average potential energy, over a period of oscillation, respectively are $K_{av}$ and $U_{av}$. Then

• A. $U_{av}=\frac{K_{av}}{2}$
• B. $U_{av}=\frac{K_{av}}{3}$
• C. $K_{av}=\frac{U_{av}}{2}$
• D. $K_{av}=U_{av}$

Answer 1

The correct option is D $K_{av}=U_{av}$

A particle is executing linear SHM. The average kinetic energy and average potential energy, over a period of oscillation, respectively are $K_{av}$ and $U_{av}$. Then

Lets consider a particle is executing linear SHM.
The average kinetic energy and average potential energy, over a period of oscillation, respectively are $K_{av}\text{and}{U}_{av}$
Ainetic energy
$K=\frac{1}{2}m{v}^2\Rightarrow K=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2\omega t$
For one cycle, the average value of
${\cos }^2\omega t=\frac{1}{2}\Rightarrow K_{av}=\frac{1}{4}m{A}^2{\omega }^2\dots \left(i\right)$
For one cycle, the average value of
${\sin }^2\omega t=\frac{1}{2}\Rightarrow U_{av}=\frac{1}{4}m{A}^2{\omega }^2\dots \left(ii\right)$
Compare equation $\left(i\right)$ and $\left(ii\right)$,
$\frac{1}{4}m{A}^2{\omega }^2=\frac{1}{4}m{A}^2{\omega }^2\Rightarrow K_{av}=U_{av}$