Question

FA point performs simple harmonic oscillation of period T and the equation of motion is given by $x = a\sin \left( {\omega t + \dfrac{\pi }{6}} \right)$. After the elapse of what fraction of the time period the velocity of the
point will be equal to half of its maximum velocity?
$\left( A \right)\dfrac{T}{3}$
$\left( B \right)\dfrac{T}{{12}}$
$\left( C \right)\dfrac{T}{8}$
$\left( D \right)\dfrac{T}{6}$

Answer 1

Hint: We know that the Simple harmonic motion is the repetitive back and fro movement through a central position or equilibrium. The time interval of each complete oscillation remains the same. Sine equation is given. Now differentiate to find the velocity. Hence from using the above statement find the time period.

Formula used:
$\dfrac{Y}{A} = \sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)$
Where $A$is the amplitude, $\lambda $ is the wavelength, $Y$ is the displacement.

Complete step by step answer:
Simple harmonic motion is the repetitive back and forth movement through a central position or equilibrium. The time interval of each complete oscillation remains the same. The force responsible for the back and fro movement is directly proportional to the distance between them. $F = - kx$ this relation is the hooke's law.
In simple harmonic motion wavelength is directly proportional to the speed and of sound and inversely proportional to the frequency of a simple harmonic motion. The particles in the medium in most of the periodic waves exhibit simple harmonic motion.
$x = a\sin \left( {\omega t + \dfrac{\pi }{6}} \right)$
Now differentiate x with respect to time t.
$\dfrac{{dx}}{{dt}} = a\omega \cos \left( {\omega t + \dfrac{\pi }{6}} \right)$
Therefore, the maximum velocity$ = a\omega $

Since $\left( {\cos \left( {\omega t + \dfrac{\pi }{6}} \right) = \dfrac{1}{2}} \right)$
We can write,
$ \Rightarrow \dfrac{{a\omega }}{2} = a\omega \cos \left( {\omega t + \dfrac{\pi }{6}} \right)$
Now we need to find the time hence,
$\omega t + \dfrac{\pi }{6} = {60^ \circ }$
We know that$\left( {\omega = \dfrac{T}{{2\pi }}} \right)$.
Substitute that in the above equation,
$t = \dfrac{\pi }{6} \times \dfrac{T}{{2\pi }}$
$t = \dfrac{T}{{12}}$

Hence, the correct answer is option (B).

Note: The particle will oscillate along the direction of the wave. The force responsible for the back and fro movement is directly proportional to the distance between them. $F = - kx$ this relation is the hook's law. The particles in the medium in most of the periodic waves exhibit simple harmonic motion.