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
B
2π/b
C
2π/b
D
2π/b

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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