Question

Consider a particle moving in simple harmonic motion according to the equation

$x=2.0cos(to\pi t+ta{n}^{-1}0.75)$

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 the acceleration have its maximum magnitude for the first time ? (c) When does the particle come to rest for the second time ?

Answer 1

$\left(a\right)x=20cos(50\pi t+ta{n}^{-1}0.75)$

$=2.0cos(50\pi t+0.643)$

$v=\frac{dx}{dt}$

$=-100\pi sin(50\pi t+0.643)=0$

$\Rightarrow sin(50\pi t+0.643)=0$

As the particle comes to rest for the 1st time

$\Rightarrow 50\pi t+0.643=\pi$

$\Rightarrow t=1.6\times {10}^{-2}sec$

(b) Acceleration,

$a=\frac{dv}{dt}$

$=-100\pi \times 50\pi cos(50\pi t+0.643)$

For maximum acceleration

$cos(50\pi t+0.643)=-1cos\pi \left(max\right)$ (so a is max)

$\Rightarrow t=1.6\times {10}^{-2}sec$

(c) When the particle comes to rest for second time,

$50\pi t+0.643=2\pi$

$\therefore t=3.6\times {10}^{-2}s$