Question

A particle of mass m is oscillating with amplitude A and angular frequency $\omega$, its average energy in one time period is ?

• A. $\frac{1}{2}m{\omega }^2{A}^2$
• B. $\frac{1}{4}m{\omega }^2{A}^2$
• C. $m{\omega }^2{A}^2$
• D. Zero

Answer 1

The correct option is A $\frac{1}{2}m{\omega }^2{A}^2$

Let $x=A\sin \omega t$

where $x$ = displacement, $a$ = amplitude, $\omega$ = angular frequency

$v$ = velocity = $\omega A\cos \omega t$

So, $K.E=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\omega t$

And potential energy $=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2\omega t$

So, total energy E = K.E+P.E

$\Rightarrow E=\frac{1}{2}m{\omega }^2{A}^2({\cos }^2\omega t+{\sin }^2\omega t)=\frac{1}{2}m{\omega }^2{A}^2$

which is constant through the motion.So average energy $=\frac{1}{2}m{\omega }^2{A}^2$