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A particle executes linear simple harmon...

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then find its time period in seconds.

Text Solution

Verified by Experts

`a=omega^(2)x,a=v,A=3cm,x=2cm`
`v=omegasqrt(A^(2)-x^(2))`
`omega^(2)x=omegasqrt(A^(2)-x^(2))`
`((2pi)/(T))2=sqrt(3^(2)-2^(2)),(4pi)/(T)=sqrt(5)`
`T=(4pi)/(sqrt(5))impliesT=5.6sec`
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  • A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is………..

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