The angular acceleration, \( \alpha \) of a body is rate of change of its angular velocity. It is expressed as rad/s². $$ a = \alpha r $$ Where, \( \alpha \) = angular acceleration
a = linear acceleration

The linear velocity, v at any point P whose distance from the central point, C is x, is given by: $$ v = \omega{\sqrt{A²-x²}} $$ At max velocity, which is attained at the central point, C: $$ x = O $$ $$ \therefore v_{midpoint} = \omega A $$ At the end or point of maximum displacement, the maximum displacement, x is equal to the amplitude of oscillation. Here, Velocity is minimum and equal to zero. $$ x = A $$ $$ V_{end} = 0 $$ Acceleration along a circular path (centripetal acceleration) is given by: $$ a = \frac{v²}{r} $$ $$ \text{Since, } v = \omega r $$ $$ a = \frac{(\omega r)²}{r} $$ $$ a = {\omega ² r} $$ Hence, at any point x : $$ a = \omega ² x $$
At end point, where x = A (acceleration is maximum) $$ a = \omega ² A $$ At mid point, where x = 0 (acceleration is minimum) $$ a = 0 $$ Note: For a body undergoing Simple Harmonic motion, the acceleration towards the center is negative (-ve) since the velocity decreases with time. Hence,

1. v = 0 at end of path of oscillation
2. a = 0 at mid point or central point, C

Example 1: A body moving with simple harmonic motion in a straight line has velocity, V and acceleration, a, when the instantaneous displacement, x in cm, from its maximum position is given by x = 2.5 sin 0.4πt, where t is in seconds. Determine the magnitude of the maximum

1. velocity
2. Acceleration

The motion of a body undergoing S.H.M is defined by the equation $$ x = Asin \omega t $$ Comparing both equations $$ A = 2.5cm = 0.025m $$ $$ \omega = 0.4π rad/s $$

1. \( \text{Linear velocity,v }= \omega A \) $$ v = 0.4π × 0.025 = 0.01π\text{ m/s} $$ $$ v = 0.03142m/s$$
2. Acceleration, a = \( \omega ² A \) $$ a = (0.4π)² × 0.025 $$ $$ = 0.004π²\text{ m/s²} $$ $$ a = 0.0395m/s² $$

Example 2: A body vibrates in simple harmonic motion with a frequency of 50Hz and an amplitude of 4cm. Find

1. The period
2. The acceleration at the middle and at the end of the path of oscillation
3. The velocities at the middle and at the end of path of oscillation
4. The velocity and acceleration at a distance of 2cm from the centre of acceleration

1. The period can be calculated as thus: $$ f = 50Hz $$ $$ T = \frac{1}{T} $$ $$ T = \frac{1}{50} = 0.02s $$ $$ \omega = 2πf $$ $$ \omega = 2π × 50 $$ $$ \omega = 100π rad/s $$
2. The acceleration at mid-point, x = 0 $$ a = \omega ² × 0 = 0 $$

At the end of the path of oscillation, x = A

3. The velocity at mid point is calculated as: $$ \text{At mid point, x = 0}$$ $$ v = \omega{\sqrt{A²-x²}} $$ $$ v = \omega A $$ $$ v = 100 \pi × 0.04 $$ $$ v = 4 \pi \text{ m/s} $$ $$ v = 12.568m/s $$ At end point, x = A $$ a = 0 $$
4. The velocity at a distance x = 2cm from the center of oscillation is given by: $$ x = 2cm = 0.02m $$ $$ A = 0.04m $$ $$ v = \omega{\sqrt{A² - x²}} $$ $$ v = 100 \pi{\sqrt{0.04²-0.02²}} $$ $$ v = 2π{\sqrt{3}} \text{ m/s} $$ $$ v = 10.88 \text{ m/s} $$ The acceleration at point x = 2cm is given as: $$ a = \omega ² x $$ $$ a = (100 \pi)² × 0.02 $$ $$ a = 200 \pi ² \text{ m/s²} $$ $$ a = 1.97 × 10^3 \text{ m/s²} $$