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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
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B
2 s
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C
4 s
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D
6 s
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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 $$\cfrac{24}{6}=4sec$$

1151436_870051_ans_152618d9fbb8410c9d133fc09c33c353.png

Physics

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