First slide

Simple harmonic motion

Question

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 $8 \sqrt{3} cm / s,$ will be

Moderate

A

$2 \sqrt{3} cm$
B

$\sqrt{3} cm$
C

1 cm
D

2 cm

Solution

At mean position velocity is maximum
i.e., $v_{\max} = ωa$ $\Rightarrow ω = \frac{v_{\max}}{a} = \frac{16}{4} = 4$
$∴ v = ω \sqrt{a^{2} - y^{2}}$ $\Rightarrow 8 \sqrt{3} = 4 \sqrt{4^{2} - y^{2}}$
$\Rightarrow 192 = 16 (16 - y^{2})$ $\Rightarrow 12 = 16 - y^{2}$ $\Rightarrow y = 2 cm .$

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