Question

The Bob of a simple pendulum oscillates with an amplitude of 20 cm. If the acceleration of the Bob is - 4y, where y is the displacement, calculate the

(I) magnitude of its maximum acceleration.

(II) period of the motion

(III) magnitude of the maximum velocity

Answer 1

Given:

$y_m=0.2\:\rm m$

$a=-4y$

The acceleration of pendulum's bob

$$a=-\omega^2y$$

Hence, the angular frequency

$$\omega=2\:\rm rad/s$$

(I) magnitude of its maximum acceleration

$$a_m=\omega^2 y_m=4*0.2=0.8\:\rm m/s^2$$

(II) period of the motion

$$T=\frac{2\pi}{\omega}=\frac{2\pi}{2}=\pi\:\rm s=3.14\: s$$

(III) magnitude of the maximum velocity

$$v_m=\omega y_m=2*0.2=0.4\:\rm m/s$$