
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
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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 \[\omega \].
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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