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Question

A particle moves such that its acceleration $$'a'$$ is given by $$a=-bx$$ where $$x$$ is the displacement from equilibrium position and $$b$$ is constant. The period of oscillation 


A
2π/b
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B
2π/b
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C
2π/b
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D
2π/b
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Solution

The correct option is A $$2\pi/\sqrt{b}$$
$$a=-\omega^2x=-bx$$
Thus, $$\omega=\sqrt b$$
Thus, $$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\sqrt b}$$

Physics

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