Maximum speed of a particle in simple harmonic motion is `v_(max)`. Then average speed of this particle in one time period is equal to

Options

A. `(v_(max))/(2)`
B. `((v_(max)))/(pi)`
C. `(piv_(max))/(2)`
D. `(2v_(max))/(pi)`

Answer

B

Explanation

Average speed =`(distance)/(time)=(4A)/(T)` (in one cycle)
`v_(max)=Aw=A(2pi)/(T)` `implies (A)/(T)=(v_max)/(2pi)` `therfore v_(avg)=4((v_(max))/(2pi))=(2V_(max))/(pi)`