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 ${{\sqrt x a} \over 2}$. The value of x is __________.
Answer (integer)
3
Solution
For a particle executes S.H.M.<br><br>$V = \omega \sqrt {{a^2} - {x^2}}$<br><br>Given, $V = {{{V_{\max }}} \over 2} = {{A\omega } \over 2}$<br><br>${{{A^2}{\omega ^2}} \over 4} = {\omega ^2}{a^2} - {\omega ^2}{x^2}$<br><br>$x = {{\sqrt 3 } \over 2}a$
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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