Oscillations

JEE Main level, numerical answer, medium

42 of 194

Question 42 of 194

A particle executes S.H.M. with amplitude AA and time period TT. The displacement of the particle when its speed is half of maximum speed is xA2.\dfrac{\sqrt{x}A}{2}. The value of xx is

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Answer: 3

V=ωA2−x2Vmax⁡=AωV=\omega \sqrt{{A}^{2}-{x}^{2}}{V}_{\max }=A\omega
Aω2=ωA2−x2\dfrac{A\omega }{2}=\omega \sqrt{{A}^{2}-{x}^{2}}
A24=A2−x2\dfrac{{A}^{2}}{4}={A}^{2}-{x}^{2}
x2=3A24{x}^{2}=\dfrac{3{A}^{2}}{4}
x=32Ax=\dfrac{\sqrt{3}}{2}A

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