1.

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

Answer» 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`


Discussion

No Comment Found

Related InterviewSolutions