Question

A particle of mass m executes shm with amplitude a and frequency v

Answer 1

Correct Question:

A particle of mass m executes shm with amplitude a and frequency $\nu$. Find it's total energy ?

Solution:

Kinetic energy of the SHM :

$\therefore \: KE = \dfrac{1}{2} m { \omega}^{2} ( {a}^{2} - {x}^{2} )$

Potential Energy of the SHM:

$\therefore \: PE= \dfrac{1}{2} m { \omega}^{2} {x}^{2}$

So, total energy :

$\therefore \: TE = PE + KE$

$\implies\: TE = \dfrac{1}{2} m { \omega}^{2} {x}^{2} + \dfrac{1}{2} m { \omega}^{2} ( {a}^{2} - {x}^{2} )$

$\implies\: TE = \dfrac{1}{2} m { \omega}^{2} {a}^{2}$

$\implies\: TE = \dfrac{1}{2} m {(2\pi \nu)}^{2} {a}^{2}$

$\implies\: TE = \dfrac{1}{2} m \times 4 {\pi}^{2} { \nu}^{2} \times {a}^{2}$

$\implies\: TE = 2m {\pi}^{2} { \nu}^{2} {a}^{2}$

So, final answer is:

$\boxed{ \bf\: TE = 2m {\pi}^{2} { \nu}^{2} {a}^{2} }$