Energy In Simple Harmonic Motion - Practice Questions & MCQ

A particle executing S.H.M. possesses two types of energy: Potential energy and Kinetic energy

Potential energy-

• This is an account of the displacement of the particle from its mean position.
• Formula-

As restoring force is given as $F=-k x$

$
\begin{aligned}
U & =-\int d w=-\int_0^x F d x=\int_0^x k x d x=\frac{1}{2} k x^2 \\
\text { using } \omega & =\sqrt{\frac{k}{m}} \text { or } k=m \omega^2
\end{aligned}
$

we get $U=\frac{1}{2} m \omega^2 x^2$
For $x=A \sin (w t)$

$
U=\frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t
$

- Potential energy maximum and equal to total energy at extreme positions

$
U_{\max }=\frac{1}{2} k A^2=\frac{1}{2} m \omega^2 A^2 \quad \text { when } x= \pm A ; \omega t=\pi / 2 ; \quad t=T / 4
$

- Potential energy is minimum at mean position
i.e $U_{\min }=0 \quad$ when $x=0 ; \omega t=0 ; t=0$
- The average value of potential energy with respect to $t$

$
\begin{aligned}
& \text { Average of } U=\frac{\int U d t}{\int d t} \\
& \because U=\frac{1}{2} k x^2 \\
& U_{\text {avg }}=\frac{\int \frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t}{\int d t}=\frac{\int \frac{1}{4} m \omega^2 A^2(1-\cos 2 \omega t) d t}{d t}=\frac{1}{4} m \omega^2 A^2
\end{aligned}
$
 

Kinetic energy-

• This is because of the velocity of the particle.
• Formula

   

$
K=\frac{1}{2} m v^2
$

or using $v=A \omega \cos \omega t$ we get $K=\frac{1}{2} m A^2 \omega^2 \cos ^2 \omega t$
And using $v=w \sqrt{A^2-x^2}$ and $k=m \omega^2$ we get

$
K . E .=\frac{1}{2} K\left(A^2-x^2\right)
$

- Kinetic energy is maximum at the mean position and equal to total energy at the mean position.

$
\text { i.e } K_{\max }=\frac{1}{2} m \omega^2 A^2 \quad \text { when } x=0 ; t=0 ; \omega t=0
$

- Kinetic energy is minimum at the extreme positions.
i.e $K_{\min }=0 \quad$ when $y=A ; t=T / 4, \omega t=\pi / 2$
- The average value of kinetic energy with respect to $t$

$
\begin{gathered}
K_{\text {avg }}=\frac{\int K d t}{\int d t} \\
K_{\text {avg }}=\frac{\int \frac{1}{2} m \omega^2 A^2 \cos ^2(\omega t)}{\int d t}=\frac{\int \frac{1}{4} m \omega^2 A^2(1+\cos 2 \omega t) d t}{d t}=\frac{1}{4} m \omega^2 A^2
\end{gathered}
$

So $K_{\text {avg }}=U_{\text {avg }}$
Total energy-
- Total mechanical energy $=$ Kinetic energy + Potential energy or $\mathrm{E}=\mathrm{K}+\mathrm{U}$

$
E=\frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2=\frac{1}{2} m \omega^2 A^2
$
 

So Total energy does not depend on position(x)  i.e. it always remains constant in SHM.

• Graph of Energy in S.H.M

At time t=0 sec, the position of the block is equal to the amplitude,