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
B
16m/sec2
C
36m/sec2
D
48m/sec2

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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