Question

Two simple harmonic motions $y_1=A\sin \omega t$ and $y_2=A\cos \omega t$ are superimposed on a particle of mass $m$. The total mechanical energy of the particle is :

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

Answer 1

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

Given
$y_1=A\sin \omega t$
$y_2=A\cos \omega t=A\sin [\frac{\pi }{2}+\omega t]$
Phase difference is
$\Delta \varphi =\frac{\pi }{2}$
Resultant - amplitude
$A_{eq}^2=A_1^2+A_2^2=2A^2$
Total mechanical energy
$E=\frac{1}{2}m{\omega }^2{A}_{eq}^2=m{\omega }^2{A}^2$