# Simple Harmonic Motion

A motion that repeats itself after a regular interval of time is called harmonic motion.
For example, backward and forward motion of a simple pendulum, up and down motion of a mass attached to a spring, etc. In harmonic motion, there is always an equilibrium position at which the body comes to rest. If the body is displaced from the equilibrium position, the system develops a restoring force in such a direction that brings it back to the equilibrium position.
In simple harmonic motion,
Restoring force ∝ displacement from equilibrium position $\text{i.e. }F ∝y$ $F = ky\text{ ___(1)}$ From Newton’s second law of motion, $F=-ma \text{ ___(2)}$ There is negative sign because it is restoring force. from (1) and (2), $ky=-ma$ $a=-\frac{k}{m}y$ $a ∝y$
Thus, a body is said to be executing simple harmonic motion if its acceleration is directly proportional to the displacement from the mean position and is directed towards the mean position.

## Uniform Circle as a Projection of SHM

​Consider a particle P is moving in a circular path with uniform angular speed. At any instant, draw PM⊥YY’. The foot M of the perpendicular is known as projection. When P moves from X to Y, its projection M moves from O to Y. And, when P moves from Y to X’, its projection M moves from Y to O. Similarly, when P moves from X’ to Y’ and Y’ to X, its projection M moves from O to Y’ and Y’ to O respectively. This back and forth motion of the projection M along the diameter is simple harmonic motion.

## Calculation of SHM

### Displacement

Consider a circular path of radius r. Let an object initially at point A reaches a point B after some time t with angular velocity ω. $∴∠AOB=θ=ωt$ In triangle OBM, $sinθ=\frac{OM}{OB}$ $sinθ=\frac{y}{r}$ $∴y=rsinωt$ This is the displacement (y) of the object executing S.H.M. It is the displacement of the particle from the mean position at that time.

### Velocity

$\text{Velocity (v)} = \text{Rate of change of y}$ $v=\frac{dy}{dt}$ $v=\frac{d(rsinωt)}{dt}$ $v=r\frac{d(sinωt)}{dωt} × \frac{d(ωt)}{dt}$ $v=r cosωt × ω$ $v=ω\sqrt{r^2cos^2ωt}$ $v=ω\sqrt{r^2(1-sin^2ωt)}$ $v=ω\sqrt{r^2-r^2sin^2ωt}$ $v=ω\sqrt{r^2-y^2}$ When the particle is at the extreme position i.e. y=r, then, $v=ω\sqrt{r^2-r^2}=0$ ∴ Velocity is zero at the extreme position. When the particle is at the equilibrium position i.e. y=0, then, $v=ω\sqrt{r^2-0}=rω$ ∴ Velocity is maximum at the equilibrium position.

### Acceleration

$Acceleration=\text{Rate of change of velocity}$ $a=\frac{dv}{dt}$ $a=\frac{d(rωcosωt)}{dt}$ $a=rω\frac{d(cosωt)}{dωt} × \frac{d(ωt)}{dt}$ $a=-rsinωt × ω$ $∴a=-ω^2y$ When the particle is at the extreme position i.e. y=r, then, $a=-ω^2r$ ∴Acceleration is maximum at the mean position. When the particle is at the mean position i.e. y=0, then, $a=-ω^2×0=0$ ∴Acceleration is zero at the mean position.

### Time Period

It is the time taken by the object to make a complete round on the circle. We know, $a=-ω^2y$ Taking magnitude only, $a=ω^2y$ $ω=\sqrt{\frac{a}{y}}$ $\frac{2 π}{T}=\sqrt{\frac{a}{y}}$ $T=2 π \sqrt{\frac{y}{a}}$ $∴T=2 π \sqrt{\frac{displacement}{acceleration}}$ This gives the time period of the particle executing S.H.M.

## Energy of a Particle Executing SHM

In simple harmonic motion, a particle possesses potential energy (U) because of its displacement from the mean position and kinetic energy (T) because of its motion.

### Potential Energy

We have, $a=-ω^2y$ Consider a body of mass m is executing SHM, then, restoring force on the body is, $F=mω^2y$ When the body moves further infinitesimally small distance dy, then the small work done is given by, $dW=Fdy$ $dW=mω^2ydy$ Total work done is given by integrating within limits from 0 to y, $∴W=\int_0^y mω^2ydy$ $W=\frac{1}{2} mω^2y^2$ This work done is equal to the potential energy (U) of the body, $∴U=\frac{1}{2} mω^2y^2$

### Kinetic Energy

Let the body have a velocity of v, then, $K.E.=\frac{1}{2}mv^2\text{ ___(a)}$ In S.H.M., $v=ω\sqrt{r^2-y^2}\text{ ___(b)}$ where, r = amplitude of the S.H.M. and y = displacement from mean position. From (a) and (b), $∴K.E. (T)=\frac{1}{2}mω^2(r^2-y^2)$

### Total Energy

$\text{Total Energy (E)}=U+T$ $E=\frac{1}{2} mω^2y^2+\frac{1}{2}mω^2(r^2-y^2)$ $E=\frac{1}{2} mω^2r^2$ This gives the total energy of the body executing S.H.M. The total energy remains constant because it does not depend upon the displacement from mean position.

### Graphical Representation

​When the body is at mean position, y = 0, then $K.E. (T)=\frac{1}{2}mω^2(r^2-0)=\frac{1}{2}mω^2r^2$ and, $∴U=\frac{1}{2} mω^2 ×0=0$ Thus, kinetic energy is maximum and potential energy is zero at the mean position. When the body is at extreme position, y = r, then $K.E. (T)=\frac{1}{2}mω^2(r^2-r^2)=0$ and, $∴U=\frac{1}{2} mω^2 r^2$ Thus, potential energy is maximum and kinetic energy is zero at the extreme position.

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