The average velocity of a particle executing SHM with an amplitude A and angular frequency `omega` during one oscillation is
A. `omegaA`
B. `(omegaA)/(2)`
C. `2omegaA//pi`
D. zero

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1 Answers

Correct Answer - D
`v_(1)=omegasqrt(A^(2)-x_(1)^(2))`
`3=omegasqrt(25-16)`
`3=3omega" "therefore omega=1`
`T=(2pi)/(omega)=(2pi)/(1)=6.28s`

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