Question

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

& A)\text{ }\dfrac{T}{3} \\\ & B)\text{ }\dfrac{T}{12} \\\ & C)\text{ }\dfrac{T}{8} \\\ & D)\ \dfrac{T}{6} \\\ \end{aligned}$$

Answer 1

We need to find the relation between the velocity of the particle undergoing simple harmonic motion and its frequency or time period. We can easily derive it from the equations of motion in Simple Harmonic oscillations of point objects.

Complete step by step answer:
Simple harmonic motion involves a period oscillation of any object about a fixed mean position. The equation of motion of a simple harmonic motion is given as -
$x=a\sin (\omega t+\phi )$ ---------(1)
where x is the position of the object under simple harmonic motion,
a is the maximum displacement from the mean position of oscillation or the amplitude of oscillation,
$\omega$ is the angular frequency of the motion,
t is an instant of time, and,
$\phi$ is the phase difference involved.
The velocity of the particle at the instant t is given by the derivation of the position equation at that instant,
i.e.,

& v=a\omega \cos (\omega t+\phi ) \\\ & \Rightarrow \text{ }{{v}_{0}}=a\omega \\\ \end{aligned}$$ ----------(2) where, $${{v}_{0}}$$ is the maximum velocity attained by the particle in the SHM. Now, let us consider the situation in the question, according to it, we are given an equation of position of the point object which can be easily compared to (1). $$x=a\sin (\omega t+\dfrac{\pi }{6})$$ The angular frequency is given as $$\omega $$, which can be easily converted to the terms of time period as – $$\begin{aligned} & \Rightarrow \text{ Frequency},\text{ }f=\dfrac{\omega }{2\pi } \\\ & \text{Time period,}T=\dfrac{1}{f} \\\ & \text{ }\Rightarrow \text{ }T=\dfrac{2\pi }{\omega }\text{ --(3)} \\\ \end{aligned}$$ Now, let us find the velocity of the given point object – $$\begin{aligned} & v=a\omega \cos (\omega t+\phi ) \\\ & v=a\omega \cos (\omega t+\dfrac{\pi }{6}) \\\ \end{aligned}$$ Now, let us substitute for the angular frequency $$\omega $$, using (3) as – $$v=a\dfrac{2\pi }{T}\cos (2\pi \dfrac{t}{T}+\dfrac{\pi }{6})\text{ --(4)}$$ From the above equation we understand that the maximum velocity is given by – $${{v}_{0}}=\dfrac{2\pi a}{T}$$ Now, we need to find the time period elapsed when $$v=\dfrac{{{v}_{0}}}{2}$$. So (4) turns out to be, $$\begin{aligned} & \\\ & \text{ }\dfrac{{{v}_{0}}}{2}=a\dfrac{2\pi }{T}\cos (2\pi \dfrac{t}{T}+\dfrac{\pi }{6})\text{ } \\\ & \Rightarrow \text{ }\dfrac{{{v}_{0}}}{2}={{v}_{0}}\cos (2\pi \dfrac{t}{T}+\dfrac{\pi }{6})\text{ } \\\ & \Rightarrow \text{ }\dfrac{1}{2}=\cos (2\pi \dfrac{t}{T}+\dfrac{\pi }{6}) \\\ & but,\text{ }\cos (2\pi \dfrac{t}{T}+\dfrac{\pi }{6})=\dfrac{1}{2}\text{ }\because \text{ }\cos (2\pi n+x)=\cos x \\\ & \Rightarrow \text{ }2\pi \dfrac{t}{T}=\text{ }\dfrac{\pi }{6} \\\ & \Rightarrow \text{ }t=\dfrac{T}{12} \\\ \end{aligned}$$ Thus, at the instant when time is one-twelfth of the time period, the velocity is the half of its maximum value. **So, the correct answer is “Option B”.** **Note:** The velocity of the particle has a phase difference of $$\dfrac{\pi }{2}$$ with the position of the particle at any instant of time. This relation describes that the maximum of the velocity and minimum of the amplitude coincides and vice-versa. This is introduced by the sine and cosine functions in the equations.