Question

A displacement of a body executing SHM is given by $x = A\sin (2\pi t + \dfrac{\pi }{3})$. The first time from $t = 0$ when the velocity is maximum is
(A) 0.33sec
(B) 0.16sec
(C) 0.25sec
(D) 0.5sec

Answer 1

Hint:- - In a simple harmonic motion (SHM) the velocity is maximum when the acceleration is minimum. In the question, we are given the displacement in SHM. First, we differentiate the given displacement equation with time to get the velocity equation. We further differentiate this velocity equation with time again to get an equation for acceleration. Now, velocity is maximum when acceleration is minimum, so we equate the acceleration equation to zero and simplify to find the time. The SI unit of time is in seconds, so $t$ is in seconds.

Complete step by step solution:
The given displacement is, $x = A\sin (2\pi t + \dfrac{\pi }{3})$
Differentiating this with respect to time (t) we get velocity (v),
$v = \dfrac{{dx}}{{dt}} = A\cos (2\pi t + \dfrac{\pi }{3}).2\pi$
$v = 2\pi A\cos (2\pi t + \dfrac{\pi }{3})$
For maximum velocity, acceleration is minimum (=0) in SHM. So,
$ \dfrac{{\partial v}}{{dt}} = 0$
$ \dfrac{{\partial v}}{{\partial t}} = - 2\pi A\sin (2\pi t + \dfrac{\pi }{3}).2\pi$
$0 = - 4{\pi ^2}A\sin (2\pi t + \dfrac{\pi }{3})$
Taking $ - 4{\pi ^2}A$to right hand side, we get,
$\sin (2\pi t + \dfrac{\pi }{3}) = 0 $
$ \sin (n\pi ) = 0 $
 $\sin (2\pi t + \dfrac{\pi }{3}) = \sin (n\pi ) $
$2\pi t + \dfrac{\pi }{3} = n\pi $
 For a SHM the maximum velocity is never negative, if we take n as 0 we get a negative value hence we take $n=1$ and do the further simplification.
$2\pi t + \dfrac{\pi }{3} = \pi $
$2\pi t = \dfrac{{2\pi }}{3} $
$t = \dfrac{1}{3} = 0.333\sec $
At time $t=0.33sec$, velocity is maximum.

Hence option (A)$0.33sec$ is the correct answer.

Additional information:Simple harmonic motion is defined as a periodic motion of a point along a straight line, such that its acceleration is always towards a fixed point in that line and is proportional to its distance from that point.

Note:The question mentions that it is simple harmonic motion so we should keep in mind that the velocity is maximum when the acceleration is minimum. The alternative method is to take the mean position as zero for maximum velocity. Further simplifying the equation by taking $\sin (n\pi ) = 0$, we get a negative value if taken as zero. So, we take n as 1 and simplify the equation to get the time for maximum velocity as 0.33sec.