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

Answer: (B)

$m{\omega }^2{A}^2$

$y=y_1+y_2=A\sin \omega t+A\cos \omega t=\sqrt{2}A\sin (\omega t+\pi /4)$
$Amplitude=\sqrt{2}A$
and total energy of the system $=\frac{1}{2}m{\omega }^2(\sqrt{2}A{)}^2=m{\omega }^2{A}^2$