Question

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

$x=2.0\cos \left(50\pi t+{\tan }^{-1}0.75\right)$

where x is in centimetre and t in second. The motion is started at t = 0.

When does the particle come to rest for the first time ?

When does the acceleration have its maximum magnitude for the first time ?

When does the particle come to rest for the second time?

Answer 1

$V=\frac{dx}{dt}=-100\sin \left(50\pi t+0.643\right)$

As particle comes to rest

$\begin{array}{c}\sin \left(50\pi t+0.643\right)=0\\ 50\pi t+0.643=\pi \\ t=1.6\times {10}^{-2}\sec \end{array}$

Maximum acceleration

$\begin{array}{c}a=\frac{dv}{dt}\\ a=-2\times 50\pi \times 50\pi \times \cos \left(50\pi t+{\tan }^{-1}\left(0.75\right)\right)\\ \cos \left(50\pi t+{\tan }^{-1}\left(0.75\right)\right)=\pi \\ t=1.6\times {10}^{-2}\sec \end{array}$

Rest for second time

$\begin{array}{c}\Rightarrow 1.6\times {10}^{-2}+\frac{T}{2}\\ 1.6\times {10}^{-2}+\frac{2\pi }{2w}\\ 1.6\times {10}^{-2}+\frac{2\pi }{\left(2\times 50\pi \right)}\\ =3.6\times {10}^{-2}\end{array}$