Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of periodic motion or oscillation where the restoring force is directly proportional to the displacement. It is characterized by a sinusoidal pattern, in which the displacement of the oscillating object, at any point in time, is proportional to the sine of the elapsed time.

Simple Harmonic Motion (SHM) is defined as a type of motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of this restoring force is always towards the mean position. The acceleration of a particle executing SHM is given by the equation: a(t) = -ω2x(t), where ω is the angular velocity of the particle.

Table of Contents

• Difference Between Periodic Oscillation and Simple Harmonic Motion
• Types of Simple Harmonic Motion
• General Terms in Simple Harmonic Motion
• Differential Equation of SHM and its Solution
• Angular Simple Harmonic Motion
• Quantitative Analysis of SHM
• Necessary Conditions for Simple Harmonic Motion
• Energy in Simple Harmonic Motion
• Geometrical Interpretation of Simple Harmonic Motion
• Horizontal Phasor Problem solving Strategy

Simple Harmonic, Periodic and Oscillatory Motion

Simple harmonic motion can be described as an oscillatory motion in which the acceleration of the particle is directly proportional to the displacement from the mean position. It is a special case of oscillatory motion.

All the Simple Harmonic Motions (SHM) are oscillatory and periodic, but not all oscillatory motions are SHM. Oscillatory motion is also known as harmonic motion of all the oscillatory motions, with the most important one being SHM.

In this type of oscillatory motion, displacement, velocity, acceleration, and force vary with respect to time in a way that can be described by either sine or cosine functions, collectively called sinusoids.

Also Read:

Simple Pendulum Concepts

Spring-Mass System

The study of Simple Harmonic Motion is very useful and forms an important tool in understanding the characteristics of sound waves, light waves and alternating currents. Any oscillatory motion which is not simple Harmonic can be expressed as a superposition of several harmonic motions of different frequencies.

Difference between Periodic, Oscillation and Simple Harmonic Motion

• Periodic Motion is a type of motion in which an object or system repeats its motion at regular intervals.

• Oscillation is a type of periodic motion in which an object or system vibrates or moves back and forth between two points.

• Simple Harmonic Motion is a type of oscillatory motion in which an object or system moves back and forth in a regular, periodic pattern. It is characterized by a restoring force that acts to bring the system back to its equilibrium position.

Periodic Motion

A motion repeats itself after an equal interval of time, such as uniform circular motion.

There is no equilibrium position.

There is no force that will restore the original state.

There is no equilibrium position that is stable.

Oscillatory Motion

• The motion of a particle moving back and forth around a mean or equilibrium position is known as an oscillatory motion.
• Oscillation is a kind of periodic motion bounded between two extreme points, such as the oscillation of a Simple Pendulum or a Spring-Mass System.
• The object will keep oscillating between two extreme points about a fixed point known as the mean position (or) equilibrium position, regardless of the path it takes.
• A restoring force will be directed towards the equilibrium position (or) mean position.
• At the mean position in an oscillatory motion, the net force on the particle is zero.
• The mean position is a stable equilibrium position.

Simple Harmonic Motion (SHM)

The path of Simple Harmonic Motion (SHM) is a constrained oscillation between two extreme points.

The path of the object must be straight.

There will be a restoring force directed towards the equilibrium position (or) mean position.

The mean position in Simple Harmonic Motion is a stable equilibrium.

Conditions for SHM:

$\begin{array}{l}\begin{array}{c}\overrightarrow{F}\propto -\overrightarrow{x}\\ \overrightarrow{a}\propto -\overrightarrow{x}\end{array}\end{array}$

Types of Simple Harmonic Motion

SHM or Simple Harmonic Motion can be divided into two categories:

Linear SHM

Angular SHM

Linear Simple Harmonic Motion

When a particle moves back and forth around a fixed point (known as the equilibrium position) along a straight line, its motion is referred to as Linear Simple Harmonic Motion.

Example: Spring-Mass System

Conditions for Linear Simple Harmonic Motion:

1. The restoring force is directly proportional to the displacement.
2. The displacement is in the opposite direction to the restoring force.
3. The acceleration is proportional to the displacement and is directed towards the origin.

The restoring force or acceleration acting on the particle should always be proportional to the displacement of the particle and directed towards the equilibrium position.

\begin{array}{l}\begin{matrix} \overrightarrow{F} \ \propto \ -\overrightarrow{x} \ \overrightarrow{a} \ \propto \ -\overrightarrow{x} \ \end{matrix}\end{array}

$\overrightarrow{x}$ - Displacement of particle from equilibrium position

$\overrightarrow{F}$ - Restoring Force

$\overrightarrow{a}$ - acceleration

Angular Simple Harmonic Motion

When a system oscillates angularly with respect to a fixed axis, then its motion is called angular simple harmonic motion.

Conditions for Executing Angular SHM:

The restoring torque acting on the particle should always be proportional to the angular displacement of the particle and directed towards the equilibrium position. Angular acceleration is also known as restoring torque.

$\theta \propto {\rm T}$ or $\theta \propto \alpha$

Τ - Torque

α Angular Acceleration

θ - Angular Displacement

Simple Harmonic Motion Key Terms

Mean Position

The net force acting on the particle is zero.

From the mean position, the force acting on the particle is

$$(\overrightarrow{F} \propto -\overrightarrow{x})$$

\begin{array}{l}\overrightarrow{a} \sim -\overrightarrow{x}\end{array}

Conditions at the Mean Position

$\begin{array}{l}\overrightarrow{F_{net}}=0\\ \overrightarrow{a}=0\end{array}$

The force acting on the particle is equal and opposite to the displacement. Therefore, this point of equilibrium will be a stable equilibrium.

Amplitude in Simple Harmonic Motion

It is the greatest possible distance of the particle from its average position.

Time Period and Frequency of Simple Harmonic Motion

The time period is the minimum amount of time after which a particle repeats its motion, or the shortest amount of time it takes to complete one oscillation.

$T=2\pi /\omega$

Frequency: The number of oscillations per second is referred to as frequency.

Frequency (f) = 1/T

Angular Frequency (ω) = 2πf = 2π/T

Phases in SHM

The phase of a vibrating particle at any instant is the state of the vibrating (or) oscillating particle regarding its displacement and direction of vibration at that particular instant.

The expression, and position of a particle as a function of time.

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

The initial phase of the particle, where $$\omega t + \Phi$$ is the phase at time t = 0, is known as the phase angle.

Phase Difference

Two particles executing simple harmonic motion with respect to the mean position have a total phase angle difference known as the phase difference. When the phase difference between two vibrating particles is an even multiple of $\pi$, they are said to be in the same phase.

$\Delta \Phi =n\pi$, where n = 0, 1, 2, 3, . . . . .

Two particles are said to be in opposite phases if the phase difference between them is an odd multiple of π.

$\Delta \Phi =(2n+1)\pi ,n=0,1,2,3,...$

Simple Harmonic Motion Equation and its Solution

Consider a particle of mass (m) executing Simple Harmonic Motion along a path x o x; the mean position at O. Let the speed of the particle be v₀ when it is at position p (at a distance n from O).

At t = 0, the particle at P is moving towards the right.

At t = t, the particle is at Q (at a distance x from O).

At a velocity (v)

\begin{array}{l}\text{The restoring force}\ \overrightarrow{F}\ \text{at Q is:}\ \overrightarrow{F} = -k\overrightarrow{Q}\end{array}

$$\overrightarrow{F} = -K\overrightarrow{x}$$

K is a positive constant.

\begin{array}{l}\Rightarrow \overrightarrow{F} = m\overrightarrow{a}\end{array}

Here, $\overrightarrow{a}$= acceleration at Q

\begin{array}{l}\Rightarrow -K\overrightarrow{x}=m\overrightarrow{a}\end{array}

$$\Rightarrow \vec{a}=-\left( \frac{K}{m} \right)\vec{x}$$

\(\frac{K}{m}={{\omega }^{2}}\)

\(\omega = \sqrt{\frac{K}{m}}\)

Since, $$(\left[ \frac{{{d}^{2}}x}{d{{t}^{2}}}=-\left( \frac{K}{m} \right)\overrightarrow{m}=-{{\omega }^{2}}\overrightarrow{x} \right]), (\Rightarrow \overrightarrow{a}=-\left( \frac{K}{m} \right)\overrightarrow{m}=-{{\omega }^{2}}\overrightarrow{x})$$

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

The differential equation for linear simple harmonic motion is $$d2x/dt2 + \omega2x = 0$$.

Solutions of Differential Equations for Simple Harmonic Motion

The solutions to the differential equation for Simple Harmonic Motion are:

The solution for the particle when it is in its mean position point (O) in figure (a) is $x=A\sin \omega t$

When the particle is at a position other than the mean position in Figure (b), $x_0=A\sin \varphi$

$x=A\sin (\omega t+\varphi )$ (when the particle at Q is in Figure (b) at any time t).

These solutions can be verified by substituting these x values in the above differential equation for the linear simple harmonic motion.

Angular Simple Harmonic Motion

A body free to rotate about an axis can make angular oscillations. For example, a photo frame or a calendar suspended from a nail on the wall. If it is slightly pushed from its mean position and released, it makes angular oscillations.

Requirements for Angular Oscillation to be Angular Simple Harmonic Motion

The body must experience a net Torque that is restored in nature. If the angle of oscillation is small, this restoring torque will be directly proportional to the angular displacement.

${\rm T}\propto -\theta$

$T=-k\theta$

${\rm T}=I\alpha$

$\alpha =-k\theta$

$$\begin{array}{l} \frac{d^2\theta}{dt^2} = -K\theta \end{array}$$

$$\frac{d^2\theta}{dt^2} = -\left(\frac{K}{I}\right)\theta = -\omega_0^2\theta$$

$$(\frac{d^2\theta}{dt^2} = -\omega_0^2\theta = 0)$$

The solution of the differential equation for an angular Simple Harmonic Motion gives the angular position of the particle with respect to time.

$$(\theta = \theta_{0} \sin\left(\omega_{0}t + \phi\right))$$

Then, angular velocity.

$$(\omega = \theta_{0} \cdot \omega_{0} \cdot \cos(\omega_{0}t + \phi))$$

${\theta }_0$ - Amplitude of the Angular SHM

Example:

• Simple Pendulum
• Seconds Pendulum
• The Physical Pendulum
• Torsional Pendulum

Quantitative Analysis of Simple Harmonic Motion

Let us consider a particle executing Simple Harmonic Motion between A and A1, passing through the mean position (or equilibrium position) O. The following is an analysis of this motion:

SHM about Position O

Position of a Particle as a Function of Time

At time t = 0, let us consider a particle that is executing Simple Harmonic Motion (SHM) and is at a distance from its equilibrium position.

Necessary Conditions for Simple Harmonic Motion

$$\overrightarrow{F} \propto -\overrightarrow{x}$$

$$(\overrightarrow{a} \propto -\overrightarrow{x})$$

\begin{array}{l} \overrightarrow{a} = -\omega^2 x \end{array}

\begin{array}{l}\overrightarrow{a} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx}\end{array}

$$\overrightarrow{a} = -\omega^2 x$$

$$(\int\limits_{0}^{v}{-{{\omega }^{2}}xdx}=\int\limits_{0}^{v}{vdv})$$

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

Some Conditions We Know:

At point A, where v = 0 and x = A, the equation (1) becomes

$$(\frac{0^2}{2} = \frac{-\omega^2 A^2}{2} + c)$$

\begin{array}{l}\frac{-{{\omega }^{2}}{{A}^{2}}}{2} = c - 0\end{array}

$$\frac{{{\omega }^{2}}{{A}^{2}}}{2} = c$$

Substitute the value of C in equation (1).

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

\begin{array}{l} \Rightarrow {{v}^{2}} = {{\omega }^{2}}\left( {{A}^{2}} - {{x}^{2}} \right) \end{array}

\begin{array}{l} \Rightarrow {{v}^{2}}=\omega^2\left(A^2-x^2\right) \end{array}

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

\begin{array}{l}\omega = \frac{v}{\sqrt{{{A}^{2}}-{{x}^{2}}}….(2)}\end{array}

Where v is the velocity of the particle executing simple harmonic motion, the instantaneous velocity is given by definition.

\begin{array}{l}v = \frac{dx}{dt} = \omega \sqrt{A^2 - x^2}\end{array}

$$\int{\frac{dx}{\sqrt{{{A}^{2}}-{{x}^{2}}}}} = \int\limits_{0}^{A}{\frac{dx}{\sqrt{{{A}^{2}}-{{x}^{2}}}}} = \int\limits_{0}^{A}{\omega dt}$$

\begin{array}{l}{{\sin }^{-1}}\left( \frac{x}{A} \right)=\omega t+\phi \\ \Rightarrow x=A\sin \left( \omega t+\phi \right) \end{array}

$$x = \sin (\omega t + \Phi) . . . . . (3)$$

The equation 3 - equation of position of a particle as a function of time.

Case 1: If at t = 0

The particle at x = x0

\begin{array}{l} \Rightarrow \sin^{-1}\left(\frac{x}{A}\right) = \omega t + \phi \end{array}

\begin{array}{l}\Rightarrow \phi = {{\sin }^{-1}}\left( \frac{{{x}_{0}}}{A} \right)\end{array}

Φ is the initial phase of the particle.

Case 2: If at t = 0

The particle at x = 0

$$(\sin^{-1}\left(\frac{O}{A}\right) = \phi)$$

Φ = 0

Case 3: If the particle is at one of its extreme positions, x = A, at t = 0

$$(\Rightarrow \sin^{-1}\left(\frac{A}{A}\right) = \phi)$$

\begin{array}{l}\Rightarrow \sin^{-1}(1) = \pi/2\end{array}

$\pi /2=\Phi$

The value of the particle at time t=0 is dependent on its position, thus making it the initial phase of the particle.

Now, if we examine the equation of the particle’s position with respect to time

$$\frac{\pi}{2} = x = A \sin (\omega t + \Phi)$$

$sin(\omega t+\Phi )$ is a periodic function with period $T=2\pi /\omega$

Which can be anything sine function or cosine function

Period of Simple Harmonic Motion

The coefficient of t is $$\omega$$.

The time period $$T = \frac{2\pi}{\omega}$$

$$\omega = \frac{2\pi}{T} = 2\pi f$$

ωt = angular frequency of SHM.

The Expression of Particle Position as a Function of Time:

\begin{array}{l}\text{Particles can be found by displacement}\ \left( \overrightarrow{x} \right), \text{velocity}\ \left( \overrightarrow{v} \right) \text{and acceleration as follows}.\end{array}

Velocity of a Particle Executing Simple Harmonic Motion

The velocity of a simple harmonic oscillator is given by v = dx/dt.

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

\begin{array}{l}v = \omega A\cos \left( \omega t+\phi \right) = \frac{d}{dt}A\sin \left( \omega t+\phi \right)\end{array}

\begin{array}{l}v = A\omega \sqrt{1-\cos^2\omega t}\end{array}

Since $$x = A \sin \omega t$$

$$(\frac{{{x}^{2}}}{{{A}^{2}}} = {{\sin }^{2}}\omega ,t)$$

\( v = A\omega \sqrt{1 - \frac{{x}^{2}}{{A}^{2}}} \)

\begin{array}{l}\Rightarrow v = \omega \sqrt{{A}^{2}-{x}^{2}}\end{array}

On Squaring Both Sides

\begin{array}{l} \Rightarrow {{v}^{2}} = \omega^2 \left(A^2 - x^2 \right) \end{array}

\begin{array}{l} \Rightarrow \frac{{{\omega }^{2}}}{{{v}^{2}}}=\left( {{x}^{2}}-{{A}^{2}} \right) \end{array}

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

This is an equation of an ellipse: $$\frac{{{v}^{2}}}{{{A}^{2}}}+\frac{{{v}^{2}}}{{{A}^{2}}{{\omega }^{2}}}=1$$

The curve between displacement and velocity of a particle executing the simple harmonic motion is an ellipse.

When $$\omega = 1$$, then the curve between v and x will be circular.

Acceleration in Simple Harmonic Motion

Hence, the expressions for displacement, velocity, and acceleration in linear simple harmonic motion are:

$\begin{array}{l}\overrightarrow{a}=\frac{dv}{dt}=\frac{d}{dt}(A\omega \cos \omega t+\varphi )\end{array}$

$\begin{array}{l}\Rightarrow \overrightarrow{a}=-{\omega }^{2}A\sin (\omega t+\varphi )\end{array}$

$\begin{array}{l}\Rightarrow \left|a\right|=-{\omega }^{2}x\end{array}$

Hence the expression for displacement, velocity and acceleration in linear simple harmonic motion are $x=Asin(\omega t+\Phi )$

$\begin{array}{l}v=A\omega \cos (\omega t+\varphi )=\omega \sqrt{A^2-x^2}\end{array}$

$\begin{array}{l}a=-A{\omega }^2\sin (\omega t+\varphi )=-{\omega }^{2}x\end{array}$

Energy in Simple Harmonic Motion (SHM)

The system that executes SHM is called the harmonic oscillator.

A particle of mass m is undergoing linear simple harmonic motion with an angular frequency of ω and an amplitude of A.

$\begin{array}{l}v=A\omega \cos (\omega t+\varphi )=\omega \sqrt{A^2-x^2}\end{array}$

$\begin{array}{l}a=-{\omega }^{2}A\sin (\omega t+\varphi )=-{\omega }^{2}x\end{array}$

$\begin{array}{l}\text{The restoring force}\left(\overrightarrow{F}\right)\text{acting on the particle is given by}\end{array}$

F = -kx, where k = m${\omega }^2$.

Kinetic Energy of a Particle in Simple Harmonic Motion

The total work done by the restoring force in displacing the particle from (x = 0) (mean position) to x = x:

When the particle has been displaced from x to x + dx, the work done by the restoring force is

dw/dx = -kx

$\begin{array}{l}=\frac{1}{2}m{v}^2[Since,v^2=A^2{\omega }^2{\cos }^2(\omega t+\varphi )]\end{array}$

$\begin{array}{l}=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2(\omega t+\varphi )\end{array}$

$\begin{array}{l}=\frac{1}{2}m{\omega }^2(A^2-x^2)\end{array}$

Therefore, the Kinetic Energy

$\begin{array}{l}=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2(\omega t+\varphi )=\frac{1}{2}m{\omega }^2(A^2-x^2)\end{array}$

Potential Energy of SHM

Potential Energy = -(Work Done by Restoring Force)

$\begin{array}{l}w=\int dw=\underset{0}{\overset{x}{\int }}-kxdx=\frac{-k{x}^2}{2}\end{array}$

$\begin{array}{l}=-\frac{m{\omega }^2{x}^2}{2}\end{array}$

$\begin{array}{l}[k=m{\omega }^2]\end{array}$

$\begin{array}{l}=-\frac{m{\omega }^2}{2}{A}^2{\sin }^2(\omega t+\varphi )\end{array}$

Potential Energy = -(work done by restoring force)

$\begin{array}{l}=\frac{m{\omega }^2{x}^2}{2}=\frac{m{\omega }^2{A}^2}{2}{\sin }^2(\omega t+\varphi )\end{array}$

Total Mechanical Energy of a Particle Executing Simple Harmonic Motion

E = KE + PE

Hence, the particle’s total energy in SHM remains constant, regardless of the instantaneous displacement.

At t = 0, when x = ±A, the relationship between Kinetic Energy, Potential Energy and time in Simple Harmonic Motion is that Kinetic Energy is 0 and Potential Energy is at its maximum.

Geometrical Interpretation of Simple Harmonic Motion

The straight line motion of the foot of the perpendicular drawn from a particle moving with uniform speed along the circumference of a circle is known as simple harmonic motion.

SHM as a Projection of Circular Motion

The particle is at position P at t = 0 and revolves along a circle with a constant angular velocity (ω). The projection of P on the diameter along the x-axis is M. At a later time (t), the particle is at Q and its projection on the diameter along the x-axis is N.

As the particle P revolves around in a circle anti-clockwise, its projection M follows it up, moving back and forth along the diameter such that the displacement of the point of projection at any time (t) is the x-component of the radius vector (A).

$x=Acos(\omega t+\Phi ).......(1)$

$y=Asin(\omega t+\Phi ).....(2)$

Thus, we can see that uniform circular motion is a combination of two linear harmonic oscillations that are perpendicular to each other.

P is under uniform circular motion, while M and N, as well as K and L, are performing simple harmonic motion about O with the same angular speed ω as that of P.

P is undergoing uniform circular motion, resulting in a centripetal acceleration along the radius vector A.

(towards the centre)

It can be broken down into two components:

$\begin{array}{l}{a}_N=A{\omega }^2{\sin }^2(\omega t+\varphi )\end{array}$

$\begin{array}{l}{a}_L=A{\omega }^2{\cos }^2(\omega t+\varphi )\end{array}$

a_N and a_L are the accelerations corresponding to the points N and L, respectively.

The foot of the projection on the x-axis in the above discussion is referred to as a horizontal phasor.

The vertical phasor, located at the foot of the perpendicular on the y-axis, executes a simple harmonic motion of amplitude A and angular frequency ω, while the horizontal phasor has a phase difference of π/2.

Problem-Solving Strategy for Horizontal Phasors

Let us assume a circle with a radius equal to the amplitude of SHM.

A particle is rotating in a circular path with the same constant velocity as that of simple harmonic motion in a clockwise direction.

The angle ϕ (phase constant) made by the particle at t = 0 with the upper vertical axis is equal to ϕ.

The horizontal component of the velocity of a particle provides information about the velocity of a particle performing the simple harmonic motion.

The horizontal component of the acceleration of a particle is equal to the acceleration of the particle performing SHM, where the acceleration is due to uniform circular acceleration centripetal and is equal to ω²A.