The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes 83cm/s, will be

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23cm

3cm

1 cm

2 cm

answer is D.

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Detailed Solution

At mean position velocity is maximum i.e., vmax=ωa ⇒ω=vmaxa=164=4∴v=ωa2−y2 ⇒83=442−y2⇒192=16 (16−y2) ⇒12=16−y2 ⇒y=2 cm.

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