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Question

For a body in $$S.H.M$$ the velocity is given by the relation $$v=\sqrt{144-16x^2}m/sec$$. The maximum acceleration is  


A
12m/sec2
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B
16m/sec2
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C
36m/sec2
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D
48m/sec2
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Solution

The correct option is C $$48 m/sec^2$$
velocity of the particle in SHM, 
$$v=\sqrt{144-16x^{2}}=4\sqrt{9-x^{2}}$$ 

comparing with $$v=\sqrt{A^{2}-x^{2}}$$
 we get $$A=3m$$ $$w=4rad/s$$

 so, maximum acceleration, 
$$a_{max}=Aw^{2}$$ 
          $$=3\times 4^{2}$$ 
$$\Rightarrow a_{max}=48m/s^{2}$$

Physics

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