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

• 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 $F=-k x$
Also, $a=\frac{F}{m} \Rightarrow a=-\frac{k}{m} x \Rightarrow a=-\omega^2 x ;$ 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.
- $\mathrm{v}=0$ at extreme position
- $\mathrm{v}=\mathrm{max}$ at mean position
- $a=0$ at mean position
- $a=\max$ at extreme position, i.e., at $x= \pm A, a= \pm \omega^2 A$
- Magnitude of maximum acceleration, $\left|a_{\max }\right|=\omega^2 A$

 

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

$
\Rightarrow \frac{d v}{d t}=-\omega^2 x \Rightarrow v \frac{d v}{d x}=-\omega^2 x \Rightarrow v d v=-\omega^2 x d x
$

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:

$
\begin{aligned}
& \int_0^v v d v=\int_A^x-\omega^2 x d x \\
& \Rightarrow\left[\frac{v^2}{2}\right]_0^v=-\omega^2\left[\frac{x^2}{2}\right]_A^x \\
& \Rightarrow v^2-0=-\omega^2\left(x^2-A^2\right) \\
& \Rightarrow v^2=\omega^2\left(A^2-x^2\right) \\
& \Rightarrow v= \pm \omega \sqrt{\left(A^2-x^2\right)} \\
& \text { At } x=0, v_{\max }= \pm \omega A
\end{aligned}
$

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

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

This can be rearranged as

$\begin{aligned} & v^2=\omega^2\left(A^2-x^2\right) \\ \Rightarrow & v^2=\omega^2 A^2-\omega^2 x^2 \\ \Rightarrow & v^2+\omega^2 x^2=\omega^2 A^2 \\ \Rightarrow & \frac{v^2}{\omega^2 A^2}+\frac{x^2}{A^2}=1\end{aligned}$

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:-
$x=A S i n(w t+\phi) ;$ where $\phi$ is initial phase or epoch and $(\omega t+\phi)$ is called as phase.
Various displacement equations:-
(1) $x=$ ASinwt $\Rightarrow$ when particle starts from mean position towards right.
(2) $x=-$ ASinwt $\Rightarrow$ when particle starts from mean position towards left.
(3) $x=A \operatorname{Coswt} \Rightarrow$ when particle starts from right extreme position towards left
(4) $x=-$ ACoswt $\Rightarrow$ when particle starts from left extreme position towards Right
2. For Velocity (v):-

$
\begin{aligned}
x & =A \operatorname{Sin}(\omega t+\phi) \\
\Rightarrow v & =\frac{d x}{d t}=A \omega \operatorname{Cos}(\omega t+\phi)=A \omega \operatorname{Sin}\left(\omega t+\phi+\frac{\pi}{2}\right)
\end{aligned}
$

3. For Acceleration:-

$
\begin{aligned}
x & =A \operatorname{Sin}(\omega t+\phi) \\
\Rightarrow v & =\frac{d x}{d t}=A \omega \operatorname{Cos}(\omega t+\phi)=A \omega \operatorname{Sin}\left(\omega t+\phi+\frac{\pi}{2}\right) \\
\Rightarrow a & =\frac{d v}{d t}=-A \omega^2 \operatorname{Sin}(\omega t+\phi)=A \omega^2 \operatorname{Sin}(\omega t+\phi+\pi)=-\omega^2 x
\end{aligned}
$

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 

• Differential equation of SHM:-

$
\begin{aligned}
& \frac{d v}{d t}=-\omega^2 x \\
& \Rightarrow \frac{d}{d t}\left(\frac{d x}{d t}\right)=-\omega^2 x \\
& \Rightarrow \frac{d^2 x}{d t^2}+\omega^2 x=0
\end{aligned}
$

If the motion of any particle satisfies this equation then that particle will do SHM.
- Different graphs in SHM

For $x=a \operatorname{Sin}(\omega t)$

Graph of displacement v/s time is given as 

Graph of velocity V/s time    

$v=a \omega \operatorname{Cos}(\omega t)$

Graph of acceleration V/s time    

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