Question

# In an angular simple harmonic motion angular amplitude of oscillation is $$\pi$$ rad and time period is $$0.4$$ sec then calculate its angular velocity at angular displacement $$\pi/2$$ rad.

A
B
C
D

Solution

## The correct option is B $$\text{42.7 rad/sec}$$$$\theta=\theta_A \sin (wt)$$$$\theta=\pi \sin \left(\dfrac{2\pi}{0.4}t\right)\rightarrow (i)$$$$w=\dfrac{d\theta}{dt}=\pi\times \dfrac{2\pi}{0.4} \cos\left(\dfrac{2\pi}{0.4}t\right)$$$$\downarrow$$                                      $$\hookrightarrow (ii)$$Angular velocityput $$\theta=\dfrac{\pi}{2}$$ in eq $$(i)$$$$\sin\left(\dfrac{2\pi}{0.4}t\right)=\dfrac{1}{2}$$$$\dfrac{2\pi}{0.4}t=\dfrac{\pi}{6}$$$$t=0.4/12$$putting $$t=0.4/12$$ in eq $$(ii)$$$$w=\dfrac{2\pi^2}{0.4}\cos \left(\dfrac{2\pi}{0.4}\times \dfrac{0.4}{12}\right)$$$$\Rightarrow \dfrac{2\pi^2}{0.4}\times \dfrac{\sqrt{3}}{2}=42.7\ rad/sec$$Option $$(B)$$Physics

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