Question

Two SHM's, $y_1=Asin\omega t\&y_2=Acos\omega t$ are superimposed on a particle of mass $m$. Total mechanical energy of 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)

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

The two $SH{M}^{\prime }s$ $y_1$ and $y_2$ are superimposed to give a new $SHM$ as $y$.

$y=y_1+y_2$

$y=Asin\left(wt\right)+Acos\left(wt\right)$

$y=\sqrt{2}A[\frac{1}{\sqrt{2}}\times sin\left(wt\right)+\frac{1}{\sqrt{2}}\times cos\left(wt\right)]$

$y=\sqrt{2}A[cos\left(\frac{\pi }{4}\right)sin\left(wt\right)+sin\left(\frac{\pi }{4}\right)cos\left(wt\right)]⟹y=\sqrt{2}A\left[sin\right(wt+\pi /4\left)\right]$

But for SHM $y=Bsin(wt+\varphi )$, total mechanical energy $T.E=\frac{1}{2}m{w}^2{B}^2$

Thus total mechanical energy for $y$, $T.E=\frac{1}{2}m{w}^2(\sqrt{2}A{)}^2$

$⟹T.E=m{w}^2{A}^2$