Question

The time period of a particle undergoing SHM is 16 sec. It starts its motion from the mean position. After 2 sec, its velocity is 0.4 m/s. the amplitude is
A . 1.44 m
B . 0.72 m
C . 2.88 m
D . 0.66 m

Answer 1

Hint: In this question, we have to find the amplitude of a particle which is in the SHM. As the particle is at the mean position so we use the formula of simple harmonic motion, differentiating it and substituting the values , we are able to get the value of amplitude and choose the correct option.

Formula Used:
Formula for the amplitude of the simple harmonic motion is
$x=A\sin (wt+\phi )$
Where x is the displacement of the particle, A is its amplitude, T is the time taken for its simple harmonic motion and T is the time taken for the specific position.

Complete step by step solution:
Given that the time period of SHM
$T=16\,s$
And the velocity $v=0.4\,m{{s}^{-1}}$
Now we use the formula of SHM
$x=A\sin (wt+\phi )$
Now weDifferentiate the above equation w.r.t x, we get
$v=Aw\cos (wt+\phi )$
Now we put the values in the above equation, we get
$v=Aw\cos \left( \dfrac{2\pi }{T}t+\phi \right)$
We know $w=\dfrac{2\pi }{T}$
$w=\dfrac{2\pi }{16}=\dfrac{\pi }{8} \\ $
$\Rightarrow v=Aw\cos \left( \dfrac{\pi }{8}t+\phi \right) \\ $
At mean position $\phi =0$. Then,
$v=A\times \dfrac{\pi }{8}\cos \left( \dfrac{\pi }{8}\times 2 \right) \\ $
$\Rightarrow 0.4=A\times \dfrac{\pi }{8}\cos \left( \dfrac{\pi }{4} \right)$
Solving the above equation, we get
$\therefore A=1.44\,m$

Thus, option A is the correct answer.

Note: We must remember the difference between the amplitude and the displacement. The displacement is the total distance covered by the object in a straight line. If we differentiate it with respect to time, then we get the velocity. It is the maximum height of the object in the simple harmonic motion.