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# Simple Harmonic Motion (S.H.M.) And Its Equation - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Equations of motions of SHM is considered one the most difficult concept.

• Simple harmonic motion is considered one of the most asked concept.

• 61 Questions around this concept.

## Solve by difficulty

A simple harmonic motion is represented by. The amplitude of the SHM is :

If a simple harmonic motion is represented by

$\dpi{100} \frac{d^{2}x}{dt^{2}}+\alpha x= 0$, its time period is

The displacement of a particle varies according to the relation $x=4\left ( \cos \pi t+\sin \pi t \right )$. The amplitude of the particle is $n\sqrt{2}$. find 'n'.

Two particles $\dpi{100} A$ and B of equal masses are suspended from two massless springs of spring constants $\dpi{100} k_{1}$ and $\dpi{100} k_{2}$ respectively. If the maximum velocities, during oscillations, are equal, the ratio of amplitudes of $\dpi{100} A$ and B is:

The function  $\dpi{100} \sin ^{2}\left ( \omega t \right )$ represents -

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## Concepts Covered - 2

Simple harmonic motion
• Periodic motion is also called as harmonic motion.

• Simple harmonic motion is the simplest form of oscillatory motion in which the particle oscillates on a straight line and the restoring force is always directed towards the mean position and it’s magnitude at any instant is directly proportional to the displacement of the particle from the mean position at that instant i.e. Restoring force α Displacement of the particle from mean position.

$F=-kx$, where x is measured from mean position

• All oscillations are not simple harmonic motions but all simple harmonic motions are oscillatory motions.

• Let’s understand SHM with the help of spring block system:

Suppose we stretch the spring to the extreme position and then release it from there.

Here we can see that acceleration is always directed towards the mean position.

And $F=-kx$

$Also,\ a=\frac{F}{m}\Rightarrow a =-\frac{k}{m}x\Rightarrow a=-\omega ^2x; where\ \omega^2 \ is\ a\ positive\ constant\ and$

$\omega=\sqrt{\frac{k}{m}}\Rightarrow k=m\omega^2$

where k is force or spring constant.

• v=0 at extreme position

• v=max at mean position

• a=0 at mean position

• $a= max\ at\ extreme\ position, i.e., at\ x=\pm A, \ a=\pm\omega^2A$

• $Magnitude\ of\ maximum\ acceleration,\left | a_{max} \right |=\omega^2A$

Equations of motions of SHM

$As\ we \ know, a=-\omega^2x$

$\Rightarrow \frac{dv}{dt}=-\omega^2x\Rightarrow v\frac{dv}{dx}=-\omega^2x\Rightarrow vdv=-\omega^2x dx$

Let the particle is released from an extreme position, i.e., at x=+A, v=0 and it becomes v when the displacement becomes x.

On integrating both sides of the above equation, we get:

$\int_{0}^{v}vdv=\int_{A}^{x}-\omega^2xdx$

$\Rightarrow \left [ \frac{v^2}{2} \right ]^{v}_0=-\omega^2\left [ \frac{x^2}{2} \right ]^{x}_A$

$\Rightarrow v^2-0=-\omega^2(x^2-A^2)$

$\Rightarrow v^2=\omega^2(A^2-x^2)$

$\Rightarrow v=\pm\omega\sqrt{(A^2-x^2)}$

$At\ x=0, v_{max}=\pm\omega A$

Note-

As the relation between velocity (v) and position (x) in SHM is given by

$v=\pm\omega\sqrt{(A^2-x^2)}$

This can be rearranged as

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

And this shows that the velocity-position graph is an ellipse (as shown in the below figure)

where

Major axis=2A

and Minor axis =$2 \omega A$

• General equation of SHM
1. For Displacement:-

$x=ASin(wt+\phi);\ where\ \phi \ is\ initial\ phase\ or \ epoch \ and\ (\omega t+\phi)\ is\ called\ as\ phase.$

Various displacement equations:-

$(1)\ x=ASin\omega t \Rightarrow \ when\ particle\ starts\ from\ mean\ position\ towards\ right.$

$(2)\ x=-ASin\omega t \Rightarrow \ when\ particle\ starts\ from\ mean\ position\ towards\ left.$

$(3) \ x=ACos\omega t \Rightarrow \ when\ particle\ starts\ from \ right \ extreme\ position\ towards \ left$

$(4)\ x=-ACos\omega t \Rightarrow \ when\ particle\ starts\ from\ left\ extreme\ position\ towards\ Right.$

1. For Velocity (v):-

$x=A Sin(\omega t+\phi )\\ \Rightarrow v=\frac{dx}{dt}=A\omega\ Cos(\omega t+\phi )=A\omega\ Sin(\omega t+\phi +\frac{\pi}{2})$

1. For Acceleration:-

$x=A Sin(\omega t+\phi )\\ \Rightarrow v=\frac{dx}{dt}=A\omega\ Cos(\omega t+\phi )=A\omega\ Sin(\omega t+\phi +\frac{\pi}{2})$

$\Rightarrow a=\frac{dv}{dt}=-A\omega^2\ Sin(\omega t+\phi )=A\omega^2\ Sin(\omega t+\phi +\pi)=-\omega^2x$

So here we can see that the phase difference between x and v is $\frac{\pi}{2}$

similarly, the phase difference between v and a is $\frac{\pi}{2}$

similarly, the phase difference between a and x is $\pi$

• Differential equation of SHM:-

$\frac{dv}{dt}=-\omega^2x$

$\Rightarrow \frac{d}{dt} \left( \frac{dx}{dt}\right )=-\omega^2x$

$\Rightarrow \frac{d^2x}{dt^2}+\omega^2x=0$

If the motion of any particle satisfies this equation then that particle will do SHM.

• Different graphs in SHM

For $x=a Sin(\omega t)$

Graph of displacement v/s time is given as

Graph of velocity V/s time

$v= a\omega\ Cos(\omega t)$

Graph of acceleration V/s time

$acceleration= -a\omega^2\ Sin(\omega t )$

## Study it with Videos

Simple harmonic motion
Equations of motions of SHM

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## Books

### Reference Books

#### Simple harmonic motion

Physics Part II Textbook for Class XI

Page No. : 344

Line : 53

#### Equations of motions of SHM

Physics Part II Textbook for Class XI

Page No. : 344

Line : 49