Question

Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos (50 πt + tan−1 0.75) where x is in centimetre and t in second. The motion is started at t = 0. (a) When does the particle come to rest for the first time? (b) When does he acceleration have its maximum magnitude for the first time? (c) When does the particle come to rest for

Answer 1

Explanation:

(a) The particle to come at rest, velocity $\mathrm{v}=\frac{d x}{d x}=0$, Thus on substituting the given value  $x=2.0 \cos \left(50 \pi t+\tan ^{-1} 0.75\right)$ we get,  $t=1.6 * 10^{-2} \mathrm{s}$ at which the particle first comes to rest.

(b) We know that $a=\frac{d v}{d t}$ and the particle to have maximum acceleration,

$\cos (50 t+0.643)=-1 \cos \pi(\max )$

Thus, on solving, we get $t=1.6 * 10^{-2} \mathrm{s}$ at which the particle acceleration will have maximum magnitude.

(c) The particle to come to rest for the second time, $50 \pi t+0.643=2 \pi$.  Therefore, on solving, we get $t=3.6 * 10^{-2} \mathrm{s}$ at which the particle comes to rest for the second time.