Answer to Question #244005 in Physics for Deby

Question

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

Answer 1

Given:

$\omega=2\pi f=2*3.14*4\:\rm Hz=25\: rad/s$

$A=0.15\:\rm m$

The equation of motion of SHO:

$$x=A\cos\omega t$$

The velocity

$$v=x'=-A\omega\sin\omega t\\=-A\omega\sqrt{1-\cos^2\omega t}=-\omega\sqrt{A^2-x^2}$$

The acceleration

$$a=v'=-A\omega^2\cos\omega t=-\omega^2x$$

Hence,

i) the maximum values of the acceleration and velocity

$$a_{\max}=\omega^2A=25^2*0.15=94.7\:\rm m/s^2$$

$$v_{\max}=\omega A=25*0.15=3.75\:\rm m/s$$

ii) the acceleration and velocity at a point 0.09m from the equilibrium

$$a=-\omega^2x=-25^2*0.09=-56.3\:\rm m/s^2$$

$$v=-25\sqrt{0.15^2-0.09^2}=-3.00\rm \: m/s$$