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Q.

A particle execute S.H.M from the mean position. Its amplitude is A, its time period is ‘T’. At what displacement, will its speed be half of its maximum speed

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a

\frac{2 A}{\sqrt{3}}

b

\frac{\sqrt{3} A}{2}

c

\frac{\sqrt{2}}{3} A

d

\frac{3A}{\sqrt{2}}

answer is A.

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Detailed Solution

V = \frac{V_{\max}}{2} = \frac{A ω}{2} \to (1)

$V = ω \sqrt{A^{2} - x^{2}} \to (2)$

$\frac{A ω}{2} = ω \sqrt{A^{2} - x^{2}}$

$\frac{A^{2}}{4} = A^{2} - x^{2}$

$x = \frac{\sqrt{3}}{2} A$

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