Answer to Question #300175 in Physics for KeKo

Question

a particle moving with SHM has a speed of 8.0ms^-1 and an acceleration of 12ms^-2 when its is 3.0m from its equilibrium position. Find : (a) the amplitude of the motion, (b) the maximum velocity, (c) the maximum acceleration.

Answer 1

Given:

$v_1=8.0\;\rm m/s$

$a_1=12.0\;\rm m/s^2$

$x_1=3.0\:\rm m$

A SHM is described by equations

$$x=A\cos\omega t\\ v=-A\omega\sin\omega t\\ a=A\omega^2\cos\omega t$$

Hence, the frequency of oscillations

$$\omega=\sqrt{a_1/x_1}=\sqrt{12.0/3.0}=2\:\rm rad/s$$

(a) The amplitude of the motion

$$A=\sqrt{x_1^2+(v_1/\omega)^2}=\sqrt{3.0^2+(8.0/2.0)^2}=5\:\rm m$$

(b) The maximum velocity

$$v_{\max}=\omega A=2.0*5.0=10.0\:\rm m/s$$

(c) The maximum acceleration

$$a_{\max}=\omega^2 A=(2.0)^2*5.0=20.0\:\rm m/s^2$$