Question

A particle is executing SHM along the x-axis given by x = A sin$\omega t.$ What is the mangnitude of the average acceleration of the partical between t = 0 and (T/4) s, where T is the time period of oscillation .

Answer 1

The particle is executing SHM as follows :-

$x=A\sin wt$

Instantaneous velocity $\left(v\right)$ of the particle is obtained as :

$v=\frac{dx}{dt}=Aw\cos wt$

as $w=\frac{2\pi }{T}$ [$T$: Time period]

$\Rightarrow v=Aw\cos \left(\frac{2\pi t}{T}\right)$

Now, at $t_1=0s$, velocity of particle is =

$v_1=Aw\cos \left(\frac{2\pi \left(0\right)}{T}\right)=Aw$

again at $t_2=T/4s$, velocity is =

$v_2=Aw\cos \left(\frac{2\pi T/4}{T}\right)=Aw\cos \left(\frac{\pi }{2}\right)=0$

Thus in time $t_1\to t_2$, average acceleration of the particle is :-

$a_{avg}=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_1}$

$=\frac{0-Aw}{\frac{T}{4}-0}$

which gives,

$a_{avg}=\frac{-4Aw}{T}$ (Ans)