First slide

Simple harmonic motion

Question

A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/sec. Its velocity will be 50 cm/sec at a distance

Moderate

A

5
B

$5 \sqrt{2}$
C

$5 \sqrt{3}$
D

$10 \sqrt{2}$

Solution

It is given $v_{\max} = 100 cm / \sec$ , a = 10 cm.
$\Rightarrow v_{\max} = aω$ $\Rightarrow ω = \frac{100}{10} = 10 rad / \sec$
Hence $v = ω \sqrt{a^{2} - y^{2}}$ $\Rightarrow 50 = 10 \sqrt{{(10)}^{2} - y^{2}}$
$\Rightarrow y = 5 \sqrt{3} cm$

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