Simple Harmonic Motion

Simple Harmonic Motion, or SHM, is a type of movement where an object repeatedly moves back and forth around a central point. The farther it moves from that point, the stronger the pull to return — like how a spring or pendulum behaves. This kind of motion is common in many physical systems and is key to understanding how waves and vibrations work in science.

1.0Types of Motion

1. Periodic Motion: Periodic motion repeats at regular time intervals, known as the time period. Example: Planetary motion around the Sun.
2. Oscillatory Motion

• Motion that moves back and forth about a fixed point.
• The fixed point is called the mean position or equilibrium position.
  Examples: Vibration of a string (e.g., Sitar)

3. Harmonic Functions

• Trigonometric functions with constant amplitude and single frequency.
• Only sin $\theta$ and cos $\theta$ functions are considered harmonic in basic form.

2.0Simple Harmonic Motion and Its Equation

SHM is an oscillatory motion where the restoring force is proportional to displacement and always directed toward the mean position.                                   

$F\propto -x\text{ or }a\propto -x$

$F=-kx$

k → Force constant of the S.H.M.

A negative sign implies that the direction of force and acceleration is towards equilibrium position and x is displacement of particles from equilibrium position.

Differential Equation of  S.H.M.

Differential equation of SHM $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$ where $\omega =\sqrt{\frac{k}{m}}$

Its solution is $\mathrm{Sin}(\omega t+\varphi )$

A=Amplitude  $\omega$=Angular Frequency  $\varphi$ =Initial Phase

3.0Linear and Angular SHM                            

4.0SHM with Phasor Diagram 

• A particle moves with uniform speed along a circle of radius A (amplitude).
• The projection (shadow) of the particle on the circle’s diameter moves back and forth along a straight line.
• This linear motion of the shadow on the diameter is called Simple Harmonic Motion.
• The particle’s position on the circle at any time t is given by the angle:
  θ=ωt where ω is the angular frequency.
• The shadow’s displacement on the diameter corresponds to the displacement in SHM.
• The direction of the particle’s velocity along the circle determines the direction of motion in SHM.

5.0Terms Associated with SHM

1. Amplitude (A): The maximum displacement of a particle performing S.H.M. from its mean position on either side is called amplitude. S.I. unit : Metre(m).
2. Range or path length of S.H.M: The distance between two extreme positions of a particle performing S.H.M. is called the path length or range of S.H.M. It is the distance equal to twice the amplitude i.e. 2A. S.I. unit : Metre(m)
3. Period (T): The time taken by a particle performing S.H.M to complete one oscillation is called the period of S.H.M. S.I. Unit : Second $T=\frac{2\pi }{\omega }$
4. Frequency (n or f): The number of oscillations of a particle performing S.H.M. in one second is called frequency. S.I. Unit : $f=\frac{1}{T}$ Hz

6.0Kinematics of SHM

1. Velocity: It is the rate of change of particle displacement with respect to time at that instant.

$x=A\mathrm{Sin}(\omega t+\varphi )$

$v=\frac{dx}{dt}=A\omega \mathrm{Cos}(\omega t+\varphi )$

At mean position (x = 0), velocity is maximum, $v_{\max }=\mp \omega A$

At extreme position (x = A), velocity is minimum, $v_{\min }=0$

Relation between v  and x ,

$\frac{x^2}{A^2}+\frac{V^2}{(A\omega {)}^2}=1$

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

2. Acceleration Analysis: It is the rate of change of particle's velocity w.r.t. time at that instant.

$x=A\mathrm{Sin}(\omega t+\varphi )$

$v=\frac{dx}{dt}=A\omega \mathrm{Cos}(\omega t+\varphi )$

$a=\frac{dv}{dt}=\frac{d}{dt}(A\omega \mathrm{Cos}(\omega t+\varphi ))$

$a=-{\omega }^{2}A\mathrm{Sin}(\omega t+\varphi )\Rightarrow a=-{\omega }^{2}x$

$a_{\max }=-{\omega }^{2}A$

7.0Graphical Representation

S.No	Graph	In form of t	In form of x	Maximum value
1.		$x=A\sin \omega t$	x =x	x=+A
2.		$v=A\omega \cos \omega t$	$v=\mp \omega \sqrt{A^2-x^2}$	$v=\mp \omega A$
3.		$a=-{\omega }^{2}A\sin \omega t$	$a=-{\omega }^{2}x$	$a=\mp {\omega }^{2}A$
4.		$F=-m{\omega }^{2}A\sin \omega t$	$F=-m{\omega }^{2}x$	$F=\mp m{\omega }^{2}A$

8.0Energy in SHM

Potential Energy (U or P.E.):

The potential energy is related to conservative force by the relation.

$U=\frac{1}{2}K{x}^2+U_0$

If at x = 0 potential energy is zero then U₀ = 0

(i) at x = 0 (Mean position) $P\cdot E_{\min }=0$

(ii) at x = ±A (extreme position) $P\cdot E_{\max }=\frac{1}{2}K{A}^2$

In Terms of Time

$x=A\mathrm{Sin}(\omega t+\varphi )$

$U=\frac{1}{2}K{A}^2{\mathrm{Sin}}^2(\omega t+\varphi )$

If initial phase $\varphi$  is zero, then

$U=\frac{1}{2}K{A}^2{\mathrm{Sin}}^2\omega t=\frac{1}{2}m{\omega }^2{A}^2{\mathrm{Sin}}^2\omega t$

Kinetic Energy (K or K.E.)

(i) In Terms of Displacement: If mass of the particle executing S.H.M. is m and its velocity is v, then kinetic energy at any instant will be

$K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2\left(A^2-x^2\right)=\frac{1}{2}k\left(A^2-x^2\right)$

(a) At mean position (x = 0)

$K\cdot E_{\max }=\frac{1}{2}k{A}^2$

(b) At extreme position (x = ±A)

$K\cdot E_{\min }=0$

(ii) In Terms of Time

$K=\frac{1}{2}m{\omega }^2{A}^2{\mathrm{Cos}}^2(\omega t+\varphi )$

If initial phase $\varphi$ is zero, then

$K=\frac{1}{2}m{\omega }^2{A}^2{\mathrm{Cos}}^2\omega t$

Total Energy in SHM

$E=P\cdot E+K\cdot E=U+K\cdot E$

(A).With Respect To Position

$E=\frac{1}{2}k{x}^2+\frac{1}{2}k\left(A^2-x^2\right)\Rightarrow E=\frac{1}{2}k{A}^2=\text{ constant }$

(B).With Respect To Time

$E=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2\omega t+\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\omega t$

$E=\frac{1}{2}m{\omega }^2{A}^2\left({\sin }^2\omega t+{\cos }^2\omega t\right)=\frac{1}{2}m{\omega }^2{A}^2=\frac{1}{2}k{A}^2=\text{ constant }$      

Note:

(1) The total energy of a particle in SHM remains constant at all times and displacements.

(2) This total energy depends on the mass, amplitude, and frequency of the vibrating particle.

Average Energy in SHM

(1) The time average of P.E. and K.E. over one cycle is

a. $<K\cdot E{>}_t=<\frac{1}{2}m{\omega }^2{A}^2{\mathrm{Cos}}^2\omega t>=\frac{1}{2}m{\omega }^2{A}^2\left(\frac{1}{2}\right)=\frac{1}{4}m{\omega }^2{A}^2=\frac{1}{4}k{A}^2$

b. $<P\cdot E{>}_t=<\frac{1}{2}m{\omega }^2{A}^2{\mathrm{Sin}}^2\omega t>=\frac{1}{2}m{\omega }^2{A}^2<{\mathrm{Sin}}^2\omega t>=\frac{1}{2}m{\omega }^2{A}^2\left(\frac{1}{2}\right)=\frac{1}{4}m{\omega }^2{A}^2=\frac{1}{4}k{A}^2$

c. $<T\cdot E{>}_t=<\frac{1}{2}m{\omega }^2{A}^2>=\frac{1}{2}m{\omega }^2{A}^2=\frac{1}{2}k{A}^2$

9.0Oscillations of Spring Block System

Spring and Restoring Force 

• When a spring is stretched or compressed slightly, a restoring force is produced.
• This force follows Hooke’s Law: $F=-kx\text{ ( }k\text{ is the spring constant.) }$
• The spring is considered massless, so the restoring force is uniform throughout.
• The spring constant k depends on the length(l) (l), radius, and material of the spring
• For a given spring: $kl=\text{ Constant }$

Spring Pendulum

$F=-kx\Rightarrow m\frac{d^{2}x}{d{t}^2}=-kx$

$\frac{d^{2}x}{d{t}^2}=-\frac{k}{m}x$

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

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

Time period; $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}$

Frequency: $n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

• The time period of a spring pendulum does not depend on gravity.
• The time period of a spring pendulum remains the same whether it oscillates vertically, horizontally, or on an inclined plane.
• Increasing the mass increases the time period of a spring pendulum:$T\propto \sqrt{m}$
• Increasing the spring constant k k decreases the time period and increases the frequency:$T\propto \frac{1}{\sqrt{k}},f\propto \sqrt{k}$
• If two masses $m_1\text{and}{m}_2$ are connected by a spring and made to oscillate the time period $T=2\pi \sqrt{\frac{\mu }{k}}$ where $\mu =\frac{m_1{m}_2}{m_1+m_2}=\text{ reduced mass }$  

• If the stretch in a vertically loaded spring is $y_0$ then for equilibrium of Mass m.

$k{y}_o=mg$

Time period; $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{y_o}{g}}$

But remember time period of spring pendulum is independent of acceleration due to gravity.

• If two particles are attached with spring in which only one is oscillating then the 

Time period $T=2\pi \sqrt{\frac{\text{ mass of oscillating particle }}{\text{ force constant }}}=2\pi \sqrt{\frac{m_1}{k}}$    

10.0Combinations of Springs

Series Combination: Here force is same in both springs $F=k_1{x}_1=k_2{x}_2$ $x=x_1+x_2$ $a=-\frac{1}{m}\left(\frac{k_1{k}_2}{k_1+k_2}\right)x$ $a=-{\omega }^{2}x$ $\omega =\sqrt{\frac{k_1{k}_2}{\left(k_1+k_2\right)m}}$ $T=2\pi \sqrt{\frac{\left(k_1+k_2\right)m}{k_1{k}_2}}=2\pi \sqrt{\frac{m}{k_{eq}}}$ $k_{eq}=\frac{k_1{k}_2}{\left(k_1+k_2\right)}$ $\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}$

Parallel combination: Extension is same for both springs but force acting will be different. $k_{1}x+k_{2}x=-ma$ $a=-\frac{\left(k_1+k_2\right)x}{m}$ $a=-{\omega }^{2}x$ $\omega =\sqrt{\frac{k_1+k_2}{m}}$ $T=2\pi \sqrt{\frac{m}{k_1+k_2}}\Rightarrow 2\pi \sqrt{\frac{m}{k_{eq}}}$ $k_{eq}=k_1+k_2$

11.0Simple Pendulum

It consists of a heavy point mass suspended by a weightless, inextensible string from a fixed support.

Note: Simple pendulum is the example of SHM but only when its angular displacement is very small.

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{\text{ length of simple pendulum }}{\text{ acceleration }}}$

12.0Cases of Effective Acceleration in a Pendulum

Periodic time of simple pendulum in reference system.

$T=2\pi \sqrt{\frac{l}{g_{eff}}}$

(A) If reference system is lift

• If velocity of life v = constant, a=0, $g_{eff}=g$ than $T=2\pi \sqrt{\frac{l}{g}}$ 

• If lift is moving upwards with acceleration a than $g_{eff}=g+a$, then Time Period is $T=2\pi \sqrt{\frac{l}{g+a}}$

• If lift is moving downwards with acceleration a than $g_{eff}=g-a$, then Time Period is $T=2\pi \sqrt{\frac{l}{g-a}}$                                   

• If lift falls downwards freely  than $g_{eff}=g-g$, then Time Period is $T=\infty$

Simple pendulum will not oscillate

Note: If a simple pendulum is shifted to the centre of earth, freely falling lift, in an artificial satellite then it will not oscillate and its time period is infinite $\left(g_{eff}=0\right)$.

(B) A simple pendulum is mounted on a moving truck

• If truck is moving with constant velocity, no pseudo force acts on the pendulum hence T remains same than time period $T=2\pi \sqrt{\frac{l}{g}}$
• If the truck accelerates forward with acceleration then a pseudo force acts in the opposite direction. So effective acceleration, $g_{eff}=\sqrt{g^2+a^2}$

$T^{\prime }=2\pi \sqrt{\frac{l}{g_{eff}}}=2\pi \sqrt{\frac{l}{\sqrt{g^2+a^2}}}\Rightarrow T^{\prime }\text{ decreases }$   

(C) If a simple pendulum of density σ oscillates in a liquid of density ρ, its time period increases compared to air and is given by:

$T=2\pi \sqrt{\frac{l}{\left[1-\frac{\rho }{\sigma }\right]g}}$

(D) If the pendulum bob carries a positive charge q and is placed in a downward uniform electric field, its time period decreases:

$T=2\pi \sqrt{\frac{l}{g+\frac{qE}{m}}}$

(E) If the pendulum bob has a positive charge q  q and oscillates in an upward electric field, its time period increases:

$T=2\pi \sqrt{\frac{l}{g-\frac{qE}{m}}}$

(F) The formula $T=2\pi \sqrt{\frac{l}{g}}$ applies only when the pendulum length l is much smaller than Earth’s radius R (l ≪ Rl ). If l is comparable to R, it is not valid.

$T=2\pi \sqrt{\frac{R}{g}}=84.6\text{ minute }\approx 1.5\text{ hour it is maximum time period }$

(G) Second’s Pendulum

If the time period of a simple pendulum is 2 seconds then it is called second’s pendulum. The second pendulum takes one second to go from one extreme position to another extreme position.

For second’s pendulum, time period, $T=2=2\pi \sqrt{\frac{l}{g}}$

So, length of second pendulum at the surface of earth  $l\approx 1\text{ metre }$

(H) When a long and short pendulum start together, they next match in phase after the short pendulum completes one extra oscillation compared to the long pendulum.

$N\sqrt{l_1}=(N+1)\sqrt{l_s}$

13.0Oscillations of General Bodies

1. S.H.M. of a liquid in U tube:

If a liquid of density   contained in a vertical U tube performs S.H.M. in its two limbs. Then time period

$T=2\pi \sqrt{\frac{L}{2g}}=2\pi \sqrt{\frac{h}{g}}$

$\text{ where }L=\text{ Total length of liquid column }$

$h=\text{ Height of undisturbed liquid in each limb }(L=2h)$

2. S.H.M. of a floating cylinder: If l l is is the length of cylinder dipping in liquid than time period 

$T=2\pi \sqrt{\frac{l}{g}}$

3. S.H.M. of a body in a tunnel dug along any chord of earth 

$T=2\pi \sqrt{\frac{R}{g}}=84.6\text{ minute }$

14.0Superposition of SHM

When a particle experiences two independent forces causing SHM, its motion is the superposition of the two simple harmonic motions.

$x_1=A_1\sin \omega t,x_2=A_2\sin (\omega t+\delta )$

$\text{ Superposition, }x=x_1+x_2$

Amplitude of Resultant SHM, $A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \delta }$