A tiny mass performs S.H.M. along a straight line with a time period of T = 0.6 sec and amplitude A = 10 cm. The mean velocity ( in m/s ) in time to displace by $\dfrac{A}{2}$ is $x$. The value of $x$ is ?

Hint: In this question, we have given a tiny mass which performs SHM. For this first we use the relation of displacement and amplitude to find the value of t then we use the formula of velocity to find out the mean velocity

Formula used:
We use the the relation of displacement and amplitude for simple harmonic motion which is
$y=a\cos (wt+\phi )$
Where x is the displacement, a in the amplitude, w is the angular velocity, t is the time and $\phi $ is the phase shift.

Complete step by step solution:
Given amplitude = 10 cm
Time period = 0.6 sec
Mean velocity to displace $\dfrac{A}{2}$ is x. We have to find the value of $x$. We know the relation of displacement and amplitude for simple harmonic motion is,
$y=a\cos (wt+\phi )$
Here $\phi $= 0 as it executes SHM from the mean position.
So, $y=a\cos wt$

Putting the values in the above equation, we get
$\dfrac{A}{2}=0.1\cos \left( \dfrac{2\pi }{0.6}\times t \right) \\ $
$\Rightarrow 0.05=\cos \left( \dfrac{10\pi t}{3} \right)$
Then $\dfrac{\pi }{3}=\left( \dfrac{10\pi t}{3} \right)$
Hence $t=0.1\sec $
Now mean velocity = $\dfrac{0.05m}{0.1\sec }$= $0.5\,m/s$

Hence the mean velocity in time to displace by $\dfrac{A}{2}$ is $0.5m/s$.

Note: Remember that all the simple harmonic motions are periodic but all periodic motions are not simple harmonic. That is because a simple harmonic motion is a type of periodic motion in which to and fro motion of an object exists about its mean position. But the mean position of the object can be different. But periodic motion is a type of motion which repeats itself after fixed intervals of time.