The amplitude of a particle executing SHM is 4 cm. At the mean position, the speed of the particle is 16 cm/s. The distance of the particle from the mean position at which the speed of the particle becomes 83 cm/s will be
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a
23 cm
b
3 cm
c
1 cm
d
2 cm
answer is D.
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Detailed Solution
At mean position, velocity is maximumi.e., vmax = ωa ⇒ ω = vmaxa = 164 = 4∴ v = ωa2-y2 ⇒ 83 = 442-y2⇒192 = 16(16-y2) ⇒ 12 = 16-y2 ⇒ y = 2cm