Energy In Simple Harmonic Motion MCQ - Practice Questions with Answers

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 $F=-k x$

$$
\begin{aligned}
& \text { So } U=-\int d w=-\int_0^x F d x=\int_0^x k x d x=\frac{1}{2} k x^2 \\
& \text { using } \omega=\sqrt{\frac{k}{m}}{ }_{\text {or }} k=m \omega^2 \\
& \text { we get } U=\frac{1}{2} m \omega^2 x^2 \\
& \text { For } x=A \sin (w t) \\
& U=\frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t
\end{aligned}
$$

- 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 the mean position

$$
\text { i.e } U_{\min }=0 \quad \text { when } x=0 ; \omega t=0 ; t=0
$$

- The average value of potential energy with respect to $t$

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

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.

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

  - 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 $t$

  $$
  \begin{aligned}
  K_{a v g} & =\frac{\int K d t}{\int d t} \\
  K_{a v g} & =\frac{\int \frac{1}{2} m \omega^2 A^2 \cos ^2(\omega t)}{\int d t}=\frac{\int \frac{1}{4} m \omega^2 A^2(1+\cos 2 \omega t) d t}{d t}=\frac{1}{4} m \omega^2 A^2 \\
  S_0 K_{a v g} & =U_{a v g}
  \end{aligned}
  $$

Total energy-

• Total mechanical energy = Kinetic energy + Potential energy  or E=K+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,