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**Hint:**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 \[\omega \].

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