A particle is executing SHM according to the equation `x = A cos omega t`. Average speed of the particle during the interval `0 le t le (pi)/(6omega)` is

Options

A. `(sqrt(3)Aomega)/(2)`
B. `(sqrt(3)Aomega)/(4)`
C. `(3 A omega)/(pi)`
D. `(3 A omega)/(pi)(2 - sqrt(3))``

Answer

D

Explanation

Average velocity `barv = (int_(0)^(t)(dx)/dt.dt)/(t)=(int_(0)^(t)dx)/(t)=(x(t)-x(0))/(t)`
`= (A(cospi//6-1))/(pi//6omega) = (3Aomega)/(pi)(sqrt(3) - 2)`
since particle doesnot change its direction in the given interval, average speed
`[bar(v)] = (3Aomega)/(pi)(2-sqrt(3))`