Question

The time period of oscillation of a S.H.O. is $\dfrac{\pi }{2}$ s. Its acceleration at a phase angle $\dfrac{\pi }{3}$ rad from extreme position is $2m{s^{ - 2}}$, what is its velocity at a displacement equals to half of its amplitude from mean position?
A) 0.707 m/s
B) 0.866 m/s
C) $\sqrt 2$m\s
D) $\sqrt 3 \,$m\s

Answer 1

Hint
Firstly, we have to calculate the angular velocity of the SHO, and then we have to find the equation for acceleration of a body in SHO. At last we will find the linear velocity of the body from acceleration and the angular displacement.

Complete step by step answer
We will start off by finding the angular velocity of the body as
$\omega {\text{ }} = {\text{ }}\dfrac{{2\pi }}{T}$
Where T is the time period and is given as $\dfrac{\pi }{2}$s in the question,
$\omega {\text{ }} = {\text{ }}\dfrac{{2\pi }}{{\pi /2}}$
$\omega {\text{ }} = {\text{ }}4{\text{ }}rad/s$
As we know that the displacement of a body is given as:
$x{\text{ }} = {\text{ }}A\sin \omega t$
Differentiating the equation with respect to time we get velocity ‘$v$’ and differentiating twice we get acceleration ‘$a$’:
$v{\text{ }} = {\text{ }}A\omega \cos \omega t \\\ a{\text{ }} = {\text{ }} - A{\omega ^2}\sin \omega t$
As the body covered an angle of $\dfrac{\pi }{2}$from the extreme position, the value of 4t becomes:
$4t{\text{ }} = {\text{ }}\dfrac{\pi }{2} + \dfrac{\pi }{3}$
Substituting the values in acceleration,
2\, = \, - A{4^2}\sin 4t \\\ 2\, = \, - A16\sin (\dfrac{\pi }{2} + \dfrac{\pi }{3}) \\\ 2\, = \, - A16\cos (\dfrac{\pi }{3}) \\\ A\, = \,\dfrac{1}{4} \\\
Now we will substitute this value of amplitude to find the linear velocity of the body
v{\text{ }} = {\text{ }}\sqrt {{A^2} - {x^2}} \\\ x\, = {\text{ }}\dfrac{A}{2} \\\ v\, = \,\sqrt {{{0.25}^2} - \dfrac{1}{{{8^2}}}} \\\ v{\text{ }} = {\text{ }}4\sqrt {\dfrac{3}{{64}}} \\\ v\, = \,\dfrac{{\sqrt 3 }}{2} \\\ v = \,0.866\,m{s^{ - 1}} \\\
Therefore, the option with the correct answer is option B.

Note
Alternatively, you can also express the displacement equation in terms of exponents as $x\, = \,\dfrac{{A\,(\,{e^{i\omega t}} - {e^{ - i\omega t}})}}{{2i}}$, and you can find the velocity and acceleration equation by differentiating this equation.