JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 21)
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
3
Explanation
For a particle executes S.H.M.
$$V = \omega \sqrt {{a^2} - {x^2}} $$
Given, $$V = {{{V_{\max }}} \over 2} = {{A\omega } \over 2}$$
$${{{A^2}{\omega ^2}} \over 4} = {\omega ^2}{a^2} - {\omega ^2}{x^2}$$
$$x = {{\sqrt 3 } \over 2}a$$
$$V = \omega \sqrt {{a^2} - {x^2}} $$
Given, $$V = {{{V_{\max }}} \over 2} = {{A\omega } \over 2}$$
$${{{A^2}{\omega ^2}} \over 4} = {\omega ^2}{a^2} - {\omega ^2}{x^2}$$
$$x = {{\sqrt 3 } \over 2}a$$
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