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.

**More on Simple Harmonic Motion**