Question

The time taken by a particle executing simple harmonic motion of period T, to move from the mean position to half the maximum displacement is
A. $\dfrac{{\rm{T}}}{2}{\rm{s}}$
B. $\dfrac{{\rm{T}}}{4}{\rm{s}}$
C. $\dfrac{{\rm{T}}}{4}{\rm{s}}$
D. $\dfrac{{\rm{T}}}{{12}}{\rm{s}}$

Answer 1

In the given question, we need to apply the concept of simple harmonic motion. The simple harmonic motion can be described by a cosine function of the displacement vector. The initial position of the particle can be taken as its equilibrium position.

Complete step by step solution:
According to the question, the particle starts from its initial position or equilibrium position at the instant, t = 0.
The displacement equation of the particle executing simple harmonic motion is written as;
$x = {A_0}{\rm{ sin}}\left( {\omega t} \right)$, where x is the instantaneous displacement, ${A_0}$is the maximum displacement of the particle from the equilibrium position, and $\omega$ is the angular frequency of oscillation.
Suppose the time taken by the particle in traveling from the equilibrium position to half the maximum displacement be ‘t’.
So, substituting all the values in the displacement equation, we get;

\;\;\;\;\dfrac{{{A_0}}}{2} = {A_0}\sin \left( {\omega t} \right)\\\ \Rightarrow \sin \left( {\omega t} \right) = \dfrac{1}{2}\\\ \Rightarrow \sin \left( {\omega t} \right) = \sin \left( {{{30}^ \circ }} \right) = \sin \left( {\dfrac{\pi }{6}} \right)\\\ \Rightarrow \omega t = \dfrac{\pi }{6} \end{array}$$ We know that, $$\omega = \dfrac{{2\pi }}{{\rm{T}}}$$ Again, substituting this expression in the above obtained value, we have; $$\begin{array}{l} \;\;\;\;\omega t = \dfrac{{2\pi }}{{\rm{T}}} \cdot t = \dfrac{\pi }{6}\\\ \Rightarrow t = \dfrac{{\rm{T}}}{{12}} \end{array}$$ **Therefore, the time taken by the particle is $$t = \dfrac{{\rm{T}}}{{12}}$$and option (D) is correct.** **Additional Information:** Simple Harmonic Motion (S.H.M.) is a particular type of periodic motion in which the force is directly proportional to the displacement of the particle from the equilibrium position. The force is always directed towards the equilibrium position of the particle. The periodic motion of spring following Hooke’s Law, the motion of a pendulum all falls under the category of simple harmonic motion. **Note:** In the given question, we cannot use $$\cos \omega t$$in place of the sine function. In this question, the initial displacement of the particle is taken to be zero, so we have to use the sine function. If the particle was released from a specific potential, then we have to use the $$\cos \omega t$$function for the displacement equation.