Question

The average kinetic energy of a simple harmonic oscillator is $2$ J and its total energy is $5$ J.Its minimum potential energy is :

• A. $1J$
• B. $1.5J$
• C. $2J$
• D. $3J$

Answer 1

The correct option is A $1J$

$T=PE+KE$
Avg $KE=\frac{KE\left(max\right)}{2}$
$\therefore KE\left(max\right)=4J$

$PE$ is minimum if $KE$ is max.
$5=PE+K{E}_{max}$

$PE=1$