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 $F=-kx$

$$\begin{array}{rl}& \text{ So }U=-\int dw=-{\int }_0^{x}Fdx={\int }_0^{x}kxdx=\frac{1}{2}k{x}^2\\ & \text{ using }\omega =\sqrt{\frac{k}{m}}{}_{\text{or }}k=m{\omega }^2\\ & \text{ we get }U=\frac{1}{2}m{\omega }^2{x}^2\\ & \text{ For }x=A\sin (wt)\\ & U=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2\omega t\end{array}$$

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

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

- Potential energy is minimum at the mean position

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

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

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

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 $K.E.=\frac{1}{2}K(A^2-x^2)$
  - Kinetic energy is maximum at the mean position and equal to total energy at the mean position.

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

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

  $$\begin{array}{rl}{K}_{avg}& =\frac{\int Kdt}{\int dt}\\ K_{avg}& =\frac{\int \frac{1}{2}m{\omega }^2{A}^2{\cos }^2(\omega t)}{\int dt}=\frac{\int \frac{1}{4}m{\omega }^2{A}^2(1+\cos 2\omega t)dt}{dt}=\frac{1}{4}m{\omega }^2{A}^2\\ S_0{K}_{avg}& =U_{avg}\end{array}$$

Total energy-

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

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

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,