Composition Of Two SHM

Periodic motion characterises all simple harmonic motions. The item oscillates in SHM, moving back and forth between its extreme and mean positions. The restoring force, which is directly proportional to the size of an object's displacement from its mean position but acts in the opposite direction as the displacement, is felt by the oscillating object during the whole oscillation process. It is possible to write it as F α -x. Cradle, swing, pendulum, guitar, bungee leaping, and other real-world instances of SHM include motions that have their restoring force opposite the displacement.

In this article, we will cover the concept of the composition of two SHM. This concept is part of Oscillations and Waves, which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, VITEEE and more. Over the last ten years of the JEE Main and NEET (from 2013 to 2023), eight and three questions have been asked on this concept respectively.

Composition of Two SHM

If a particle is acted upon by two forces such that each force can produce SHM, then the resultant motion of the particle is a combination of SHM.

Composition of two SHM in the same direction

Let a force $F_1$ produces an SHM of amplitude $A_1$ whose equation is given by:

$x_1=A_1 \sin \omega t$

Let another force $F_2$ produce an SHM of amplitude $A_2$ whose equation is given by:

$x_2=A \sin (\omega t+\phi)$

Now if force $F_1 \text { and } F_2$ is acted on the particle in the same direction then the resultant amplitude of the combination of SHM's is given by

$A=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cdot \cos \phi}$

$A_1 \text { andA }_2$ are the amplitude of two SHM's. $\phi$ is phase difference.

Note: Here the frequency of each SHM's are the same

And the resulting phase is given by

$\phi^{\prime}=\tan ^{-1}\left(\frac{A_2 \sin \phi}{A_1+A_2 \cos \phi}\right)$

Composition of SHM in Perpendicular Direction:

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Let a force F₁ on a particle produces an SHM given by

$x=A \sin \omega t$

and a force F₂ alone produces an SHM given by

$x=A \sin (\omega t+\phi)$

Both the force F₁ and F₂ acting perpendicular on the particle will produce an SHM whose resultant is given by:

$\frac{x^2}{A_1{ }^2}+\frac{y_2{ }^2}{A_2{ }^2}-\frac{2 x y \cos \phi}{A_1 A_2}=\sin ^2 \phi$

The above equation is the general equation of an ellipse. That is two forces acting perpendicular on a particle execute SHM along an elliptical path.

$\text { When } \phi=0 \text { resultant equation is given by }$

$ y=\frac{A_2}{A_1} \cdot x$

It is a straight line with a slope

$\frac{A_2}{A_1}$ represented by the below figure

When $\phi=\pi$ resultant equation
$
y=\frac{-A_2}{A_1} \cdot x
$

which is represented by below straight line with slope $\frac{-A_2}{A_1}$

When $\phi=\frac{\pi}{2}$ resultant equation
$
\frac{x^2}{A_1{ }^2}+\frac{y^2}{A_2{ }^2}=1
$

It represents a normal ellipse
${ }_{\text {if }} A_1=A_2$ and $\phi=\frac{\pi}{2}$ then it represents a circle.

Solved Examples Based on the Composition Of Two SHM

Example 1: The SHM of a particle is given by the equation $y=3 \sin \omega t+4 \cos \omega t$. The amplitude is:

1) 5

2) 1

3) 7

4) 12

Solution:

Resultant Amplitude of Two SHM -

$
A=\sqrt{A_1{ }^2+A_2^2+2 A_1 A_2 \cdot \cos \phi}
$
$A_1$ and $A_2$ are amplitude of two SHMs. $\phi$ is phase difference
$
\text { Resultant Amplitude }=\sqrt{3^2+4^2}=5
$

Hence, the answer is the option (1).

Example 2: The motion of a particle varies with time according to the relation $y=a(\sin \omega t+\cos \omega t)$, then-

1) The motion is oscillatory t but not SHM
2) The motion is SHM with amplitude $a$
3) The motion is SHM with amplitude $a \sqrt{2}$
4) The motion is SHM with amplitude $2 a$

Solution:

Both SHMs are along the same direction and of the same frequency.

$\begin{aligned}
& y=a(\cos \omega t+\sin \omega t)=a \sqrt{2}\left[\frac{1}{\sqrt{2}} \cos \omega t+\frac{1}{\sqrt{2}} \sin \omega t\right] \\
\Rightarrow & y=a \sqrt{2}\left[\sin 45^{\circ} \cos \omega t+\cos 45^{\circ} \sin \omega t\right]=a \sqrt{2} \sin \left(\omega t+45^{\circ}\right) \\
\Rightarrow & \text { Amplitude }=a \sqrt{2}
\end{aligned}$

Hence, the answer is the option (3).

Example 3: A particle executing simple harmonic motion along $y$-axis has its motion described by the equation $y=A \sin (\omega t)+B$. The amplitude of the simple harmonic motion is:

1) $A$
2) $\mathrm{B}$
3) $A+B$
4) $\sqrt{A+B}$

Solution:

The amplitude is the maximum displacement from the mean position.

This question is based on the concept of shifting of mean position.

At mean position

$ Y=y-B=0 \Rightarrow y=B$

and the equation is $Y=y-B=A \operatorname{Sin}(\omega t)$

So the Amplitude is A.

Hence, the answer is the option (1),

Example 4: A simple harmonic oscillator of angular frequency 2 rad s^-1 is acted upon by an external force F=sint N. If the oscillator is at rest in its equilibrium position at t=0, its position at later times is proportional to:

1) $\sin t+\frac{1}{2} \sin 2 t$
2) $\sin t+\frac{1}{2} \cos 2 t$
3) $\cos t-\frac{1}{2} \sin 2 t$
4) $\sin t-\frac{1}{2} \sin 2 t$

Solution:

From the equation of motion, we have

$\begin{aligned}
& F(t)=k x+m \ddot{x} \ldots \ldots (1) \\
& \frac{F(t)}{m}=w_0^2 x+\ddot{x}_{\ldots \ldots (2) }
\end{aligned}$

The general solution of equation (2) consists of a sum of two parts,

The first part is the solution let's say x=P(t) which satisfies equation (2), is called a particular solution.

The second part is the solution let's say x=S(t) which satisfies equation (2) with F(t)=0, is called a specific solution.

for x=P(t)

$\frac{\mathrm{d}^2 \mathrm{P}(\mathrm{t})}{\mathrm{dt}^2}+\omega_0^2 \mathrm{P}(\mathrm{t})=\frac{\mathrm{F}(\mathrm{t})}{\mathrm{m}}$

We try a solution of type $\mathbf{P}(\mathbf{t})=\mathbf{A}_1 \sin \omega t$ whose frequency is the same as of forcing frequency which is equal to 1, and $ \omega_0=2 \mathrm{rad} / \mathrm{s}$

So

$\begin{aligned}
& -\mathbf{A}_1 \sin (t)+2^2 \mathbf{A}_1 \sin (t)=\frac{\sin t}{\mathrm{~m}} \\
& \Rightarrow \mathbf{A}_1=\frac{\frac{1}{\mathrm{~m}}}{4-1}=\frac{1}{3 \mathrm{~m}}
\end{aligned}$

and the specific solution is given by

$\frac{\mathrm{d}^2 \mathrm{~S}(\mathrm{t})}{\mathrm{dt}^2}+\omega_0^2 \mathrm{~S}(\mathrm{t})=0$

For which the solution is given as of SHM

$ \text { i.e } \mathbf{S}(\mathbf{t})=\mathbf{A}_2 \sin \left(\omega_0 \mathbf{t}-\phi\right)=\mathbf{A}_2 \sin (2 \mathbf{t}-\phi)$

$\text { where } A_2 \text { and } \phi \text { are determined by initial conditions }$

Now The general solution is given as

$
\mathbf{x}(\mathbf{t})=\mathbf{P}(\mathbf{t})+\mathbf{S}(\mathbf{t})=\frac{1}{3 \mathbf{M}} \sin (\mathbf{t})+\mathbf{A}_2 \sin (\mathbf{2 t}-\phi)
$

Given $\quad x(t)=0$ at $t=0$ and $\left(\frac{d x}{d t}\right)_{t=0}=0$

So using x(t)=0 at t=0

$\mathbf{0}=\mathbf{0}+\mathbf{A}_2 \sin (\mathbf{0}-\phi) \Rightarrow \phi=2 \mathbf{k} \pi$ where $\mathbf{k}$ is an integer
$
\text { As } \frac{d x}{d t}=\frac{1}{3 M} \cos (t)+A_2 \times 2 \times \cos (2 t-2 k \pi)
$

Now using $\left(\frac{d x}{d t}\right)_{t=0}=0$

we get

$\begin{aligned}
& \Rightarrow \frac{1}{3 \mathrm{M}}+\mathrm{A}_2 \times 2=0 \\
& \Rightarrow \mathrm{A}_2=-\frac{1}{2} * \frac{1}{3 \mathrm{M}}
\end{aligned}$

$\text { substituting the value of } A_2 \text { in the general solution of } \mathrm{x}(\mathrm{t})$

$\mathrm{x}(\mathrm{t})=\frac{1}{3 \mathrm{M}} \sin (\mathrm{t})-\left(\frac{1}{2} * \frac{1}{3 \mathrm{M}} \sin (2 \mathrm{t}-2 \mathrm{k} \pi)\right)$

taking k=0

$\Rightarrow x(t)=\frac{1}{3 M}\left(\sin (t)-\frac{1}{2} \sin (2 t)\right)$

Hence, the answer is the option (4).

Example 5: A particle executes simple harmonic motion and is located at $\mathrm{x}=\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ at times $t_0, 2 t_0, 3 t_0$ respectively. The frequency of the oscillation is :

$\begin{aligned}
& \frac{1}{2 \pi t_0} \cos ^{-1}\left(\frac{a+c}{2 b}\right) \\
& \text { 2) } \frac{1}{2 \pi t_0} \cos ^{-1}\left(\frac{a+b}{2 c}\right) \\
& \text { 3) } \frac{1}{2 \pi t_0} \cos ^{-1}\left(\frac{2 a+3 c}{b}\right) \\
& \text { 4) } \frac{1}{2 \pi t_0} \cos ^{-1}\left(\frac{a+2 b}{3 c}\right)
\end{aligned}$

Solution:

$\begin{aligned}
& x=A \sin (w t+\phi) \\
& \text { let } \phi=0 ; x=A \sin w t \\
& \text { then } a=A \sin w t_0 \\
& b=A \sin 2 w t_0 \\
& c=A \sin 3 w t_0 \\
& a+c=A\left(\sin w t_0+\sin 3 w t_0\right)=2 A \sin 2 w t_0 \cos \left(w t_0\right) \\
& a+c=b\left(2 \cos w t_0\right) \\
& \therefore w=\frac{1}{t_0} \cos ^{-1}\left(\frac{a+c}{2 b}\right) \Rightarrow f=\frac{1}{2 \pi t_0} \cos ^{-1}\left(\frac{a+c}{2 b}\right)
\end{aligned}$

Hence, the answer is the option 1.

Summary

Adding two oscillatory motions of two Simple Harmonic Motions (SHM) with the possibility of different amplitudes, frequencies and phases leads to a compound motion. The resultant would be found by adding their individual displacements using vector addition provided they are along one line. In addition, if they have similar frequencies, the resultant motion will be SHM but with a different amplitude and phase.

Frequently Asked Questions (FAQs):

Q 1. What is Simple harmonic motion?

Ans: 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 its 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 the mean position.

Q 2: Give the example of periodic motion.

Ans: Circular motion with uniform speed.

Q 3: What is Osillation?

Ans: An Oscillation is a special type of periodic motion in which a particle moves to and fro about a fixed point called the mean position of the particle.

Q 4: Which of the following is a necessary and sufficient condition for SHM?

Ans: Mean Position: A position during oscillation where the particle is at the equilibrium position, i.e. net force on the particle at this position is zero.