Question

Maximum velocity ini SHM is `v_(m)`. The average velocity during motion from one extreme point to the other extreme point will be
A. `(4)/(pi)V_(max)`
B. `(pi)/(4)V_(max)`
C. `(2)/(pi)V_(max)`
D. `(pi)/(2)V_(max)`

Answer 1

Correct Answer - C
`v_(max)=Aomega=(2piA)/(T)` and A, Amplitue, T, Period
`rArr" "(A)/(T)=(V_(max))/(2pi)`
`"Average velocity"=("Total displacement")/("Total time")=(2A)/(T//2)=(4A)/(T)`
But `((A)/(T)=(V_(max))/(2pi))`
`therefore` Average velocity `= 4 ((V_(max))/(2pi))=(2)/(pi)V_(max)`