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The maximum velocity a particle, executing simple harmonic motion with an amplitude 7 mm, 4.4 m//s. The period of oscillation is.A. 0.01 sB. 10 sC. 0.1 sD. 100 s

Answer» Correct Answer - B
The maximum velocity of a particle performing SHM is given by `v = A omega`, where A is the amplitude and `omega` is the angular frequency of oscillation.
`because" "v = A omega`
`rArr" "4.4 = 7 xx (2pi)/(T)`
`therefore" "T = (2 pi xx 7)/(4.4) = 9.99 sec = 10 sec`


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