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

It is given that a particle executes S.H.M.
Equation of S.H.M. of the particle:
x = 2.0 cos (50$\mathrm{\pi }$t + tan⁻¹0.75)
= 2.0 cos (50$\mathrm{\pi }$t + 0.643)

(a) Velocity of the particle is given by,
$v=\frac{dx}{dt}$
v = −100$\mathrm{\pi }$ sin (50$\mathrm{\pi }$t + 0.643)

As the particle comes to rest, its velocity becomes be zero.
⇒​ v = −100$\mathrm{\pi }$ sin (50$\mathrm{\pi }$t + 0.643) = 0
⇒ sin (50$\mathrm{\pi }$t + 0.643) = 0 = sin $\mathrm{\pi }$

When the particle initially comes to rest,
50$\mathrm{\pi }$t + 0.643 = $\pi$
⇒ t = 1.6 × 10⁻² s

(b) Acceleration is given by,
$$\begin{gathered} a=\frac{dv}{dt} \\ =-100\mathrm{\pi }\times 50\pi \cos \left(50\mathrm{\pi }t+0.643\right) \end{gathered}$$

For maximum acceleration:
cos (50$\mathrm{\pi }$t + 0.643) = −1 = cos $\mathrm{\pi }$ (max) (so that a is max)
⇒ t = 1.6 × 10⁻² s

(c) When the particle comes to rest for the second time, the time is given as,
50$\mathrm{\pi }$t + 0.643 = 2$\mathrm{\pi }$
⇒ ​ t = 3.6 × 10⁻² s