Question

A particle moves with simple harmonic motion along x-axis. At time t and 2t, its positions are given by $x = a$ and $x = b$ respectively from equilibrium positions. Find the time period of oscillations.

Answer 1

Express the displacements of the particle using a wave equation. Rearrange these two equations of displacement and determine the value of angular frequency. Then use the relation between angular frequency and period of the wave.

Formula used:
$\Rightarrow\omega = \dfrac{{2\pi }}{T}$
Here, $\omega$ is the angular frequency and T is the period of the wave.

Complete step by step answer:
The displacement of the particle from the mean position is given by the wave equation,
$\Rightarrow x = A\sin \omega t$
Here, A is the amplitude of the wave, $\omega$ is the angular frequency and t is the time.
Write the displacements of the wave at time t and 2t as follows,
$\Rightarrow a = A\sin \omega t$ …… (1)
$\Rightarrow b = A\sin \left( {2\omega t} \right)$ …… (2)
Divide equation (2) by equation (1).
$\Rightarrow\dfrac{b}{a} = \dfrac{{\sin \left( {2\omega t} \right)}}{{\sin \omega t}}$
Use the identity, $\sin 2\theta = 2\sin \theta \cos \theta$ to rewrite the above equation as follows,
$\Rightarrow\dfrac{b}{a} = \dfrac{{2\left( {\sin \omega t} \right)\left( {\cos \omega t} \right)}}{{\sin \omega t}}$
$\Rightarrow \dfrac{b}{a} = 2\cos \omega t$
Rewrite the above equation for $\omega t$.
$\Rightarrow\omega t = {\cos ^{ - 1}}\left( {\dfrac{b}{{2a}}} \right)$ …… (3)
The angular frequency of the wave is expressed as,
$\Rightarrow\omega = \dfrac{{2\pi }}{T}$
Here, T is the period of the wave.
Therefore, equation (3) becomes,
$\Rightarrow\dfrac{{2\pi t}}{T} = {\cos ^{ - 1}}\left( {\dfrac{b}{{2a}}} \right)$
$\Rightarrow T = \dfrac{{2\pi t}}{{{{\cos }^{ - 1}}\left( {\dfrac{b}{{2a}}} \right)}}$
This is the period of the oscillations of the given wave.

Note: In formula $\sin 2\theta = 2\sin \theta \cos \theta$, the angle $\theta$ is considered as $\omega t$ and not just $\omega$.