First slide

Simple harmonic motion

Question

Velocity at mean position of a particle executing S.H.M. is v, they velocity of the particle at a distance equal to half of the amplitude

Moderate

A

4v
B

2v
C

$\frac{\sqrt{3}}{2} v$
D

$\frac{\sqrt{3}}{4} v$

Solution

Velocity in mean position $v = aω,$ velocity at a distance of half amplitude.
$v^{'} = ω \sqrt{a^{2} - y^{2}}$ $= ω \sqrt{a^{2} - \frac{a^{2}}{4}}$ $= \frac{\sqrt{3}}{2} aω = \frac{\sqrt{3}}{2} v$

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