QJEEM21Ph3da9: JEE Main 2021 Physics - Class 11 JEEM21Ph3da9 Solution: A particle executes S.H.M. with amplitude 'a', and time period…

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A particle executes S.H.M. with amplitude 'a', and time period 'T'. The displacement of the particle when its speed is half of maximum speed is

\frac{\sqrt{x} a}{2}

. The value of x is __________.

Accepted Answer

For a particle executes S.H.M.

V = ω \sqrt{a^{2} - x^{2}}

Given,

V = \frac{V_{max}}{2} = \frac{A ω}{2}

\frac{A^{2} ω^{2}}{4} = ω^{2} a^{2} - ω^{2} x^{2}

x = \frac{\sqrt{3}}{2} a

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