Question

# How long after the beginning of motion is the displacement of a harmonically oscillating particle equal to one half its amplitude if the period is $$24\ s$$ and particle starts from rest

A
12 s
B
2 s
C
4 s
D
6 s

Solution

## The correct option is C $$4\ s$$Let's draw the phasor diagram$$\cos\theta=\cfrac{\cfrac{A}{2}}{A}=\cfrac{1}{2}\\ \theta 60°$$On rotation of $$960°$$ it takes $$24$$ sec the for $$60°$$ rotation it will take $$\cfrac{24}{6}=4sec$$Physics

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