# 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 [Helical 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 $OBM$, $\sin θ=\frac{OM}{OB}$ $\sin θ=\frac{y}{r}$ $∴y=r\sin ωt$ 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

$\text{Velocity (v)} = \text{Rate of change of y}$ $v=\frac{dy}{dt}$ $v=\frac{d(r\sin ωt)}{dt}$ $v=r\frac{d(\sin ωt)}{dωt} × \frac{d(ωt)}{dt}$ $v=r \cos ωt × ω$ $v=ω\sqrt{r^2\cos^2ωt}$ $v=ω\sqrt{r^2(1-\sin^2ωt)}$ $v=ω\sqrt{r^2-r^2\sin^2ωt}$ $v=ω\sqrt{r^2-y^2}$ 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

$Acceleration=\text{Rate of change of velocity}$ $a=\frac{dv}{dt}$ $a=\frac{d(rω\cos ωt)}{dt}$ $a=rω\frac{d(\cos ωt)}{dωt} × \frac{d(ωt)}{dt}$ $a=-r\sin ωt × ω$ $∴a=-ω^2y$ 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, $a=ω^2y$ $ω=\sqrt{\frac{a}{y}}$ $\frac{2 π}{T}=\sqrt{\frac{a}{y}}$ $T=2 π \sqrt{\frac{y}{a}}$ $∴T=2 π \sqrt{\frac{\text{displacement}}{\text{acceleration}}}$ 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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