1.

A particle executing SHM has a frequency of 10 Hz when it crosses its equilibrium position with a velocity of `2pi m//s`. Then the amplitude of vibration isA. 0.1 mB. 0.2 mC. 0.4 mD. 1 m

Answer» Correct Answer - A
`v_(m)=Aomega`
`therefore A=(v_(m))/(2pin)=(2pi)/(2pixx10)=0.1m`


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