17.3 Uniform Circular Motion and SHM | Simple Harmonic Motion | Physics 12

Simple Harmonic Motion (SHM) can be understood geometrically by analysing the projection of Uniform Circular Motion (UCM) onto a diameter.

The Reference Circle

Consider a particle $P$ moving with constant speed around a circle of radius $x_{0}$ (the reference circle). The radius $x_{0}$ equals the amplitude of the corresponding SHM. If $P$ completes one revolution in time $T$ , its angular frequency is:

$ω = \frac{2 π}{T}$

The projection of $P$ onto the x-axis (a diameter) moves back and forth between $- x_{0}$ and $+ x_{0}$ , executing SHM.

Displacement

If at $t = 0$ the particle is at the rightmost point of the circle, the angle swept in time $t$ is $θ = ω t$ . The x-projection gives the instantaneous displacement:

$x = x_{0} cos (ω t)$

where $x_{0}$ is the amplitude and $ω$ is the angular frequency.

Velocity

The particle $P$ moves with tangential speed $v_{0} = ω x_{0}$ . The x-component of this velocity gives the instantaneous velocity of the SHM projection:

$v = - v_{0} sin (ω t) = - ω x_{0} sin (ω t)$

This can also be written as:

$v = v_{0} cos (ω t + \frac{π}{2})$

In terms of displacement $x$ , using $sin^{2} (ω t) + cos^{2} (ω t) = 1$ :

$v = \pm ω x_{0}^{2} - x^{2}$

Extreme values of velocity:

At the mean position ( $x = 0$ ): $v = \pm ω x_{0}$ (maximum)
At the extreme positions ( $x = \pm x_{0}$ ): $v = 0$

Acceleration

The centripetal acceleration of $P$ points toward the centre of the circle with magnitude $ω^{2} x_{0}$ . Its x-component gives the instantaneous acceleration of the SHM projection:

$a = - ω^{2} x_{0} cos (ω t) = - ω^{2} x$

The negative sign confirms that acceleration is always directed toward the mean position (restoring in nature).

Extreme values of acceleration:

At the mean position ( $x = 0$ ): $a = 0$
At the extreme positions ( $x = \pm x_{0}$ ): $∣ a ∣ = ω^{2} x_{0}$ (maximum)

Summary Table

Quantity	At Mean Position ( $x = 0$ )	At Extreme ( $x = \pm x_{0}$ )
Displacement $x$	0	$\pm x_{0}$
Velocity $v$	$\pm ω x_{0}$ (max)	0
Acceleration $a$	0	$\mp ω^{2} x_{0}$ (max magnitude)

Key Equations

$x = x_{0} cos (ω t)$ $v = v_{0} cos (ω t) (where v_{0} = ω x_{0}, starting from extreme)$ $v = \pm ω x_{0}^{2} - x^{2}$ $a = - ω^{2} x$

Quantity

At Mean Position (

x = 0

)

At Extreme (

x = \pm x_{0}

)

Displacement

x

\pm x_{0}

Velocity

v

\pm ω x_{0}

(max)

Acceleration

a

\mp ω^{2} x_{0}

(max magnitude)