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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 at
mean position. Then,
$$x= A\, sin(\omega t)$$ is the expression
of 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, $$\dfrac{2\pi t}{T}= cos^{-1}\left ( \dfrac{b}{2a} \right )$$
$$\Rightarrow T= \dfrac{2\pi t}{cos^{-1}\left ( \dfrac{b}{2a} \right )}$$

1112445_1115135_ans_ccd4a4fcbacd465186427107d782dd50.jpg

Physics

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