Answer to Question #224738 in Mechanics | Relativity for Nafiz

Question

A particle executes SHM of amplitude 5m when the particle is 3m from its mean position, its
acceleration is found to be 48m/s2
. Find (i) velocity (ii) time period (iii) Maximum velocity

Answer 1

$v=w\sqrt{a^2-x^2}$

$$v=w\sqrt{5^2-3^2}=w\sqrt{25-9}=w\sqrt16=4w$$

$a=w^2x=w^2\times3$

$48=3w^2\\w^2=16$

$w=4rad/sec$

$T=\frac{2\pi}{4}=1.57sec$

V=aw

$v_{max}=5\times4=20$