Question

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

Solution

## If initially particle is atmean position. Then,$$x= A\, sin(\omega t)$$ is the expressionof its displacement from mean. So,$$a= A\, sin (\omega t)$$ ____ (i)$$b= A\, sin (2\omega t)$$ ____ (ii)Dividing (ii) by (i), we get$$\dfrac{sin (2 cos)}{sin (\omega t)}=\dfrac{b}{a}$$$$\dfrac{2sin(\omega t)cos (\omega t)}{sin (\omega t)}=\dfrac{b}{a}$$$$\Rightarrow cos(\omega t)=\dfrac{b}{2a}$$$$\Rightarrow \omega t= cos^{-1}\left ( \dfrac{b}{2a} \right )$$As $$\omega = 2\pi f\Rightarrow \omega =\dfrac{2\pi }{1}$$So, $$\dfrac{2\pi t}{T}= cos^{-1}\left ( \dfrac{b}{2a} \right )$$$$\Rightarrow T= \dfrac{2\pi t}{cos^{-1}\left ( \dfrac{b}{2a} \right )}$$Physics

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