Question

A body vibrating with simple harmonic motion has a frequency of 4Hz and an amplitude of 0.15m. Calculate:
a) the maximum values of the acceleration and velocity
b) the acceleration and velocity at a point 0.09m from the equilibrium position of motion

Answer 1

Given:

$f=4\:\rm Hz$

$A=0.15\:\rm m$

(a) the maximum velocity

$$v_m=\omega A=2\pi f A=2\pi *4*0.15=3.8\:\rm m/s$$

the maximum acceleration

$$a_m=\omega^2 A=4\pi^2 f^2 A=(2\pi *4)^2*0.15=95\:\rm m/s^2$$

(b)

$$v=\sqrt{\omega^2(A^2-x^2)}\\ =\sqrt{631(0.15^2-0.09^2)}=3.0\:\rm m/s$$

$$a=\omega^2 x=631*0.09=57\:\rm m/s^2$$