Frequency of a particle executing SHM is 10 Hz. The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is (g=10m/`s^(2)`

Options

A. `2pi(m)/(s)`
B. `pi(m)/(s)`
C. `(1)/(pi)(m)/(s)`
D. `(1)/(2pi)/(m)/(s)`

Answer

D

Explanation

Mean position of the particle is `(mg)/(k)` distance below the unstretched position of spring. Therefore aplitude of oscillation `A=(mg)/(k)`
`omega=sqrt((k)/(m))=2pif=20pi(f=10Hz)`
`(m)/(k)=(1)/(400pi^2)`
`v_(max)=Aomega=(g)/(400pi^2)xx20pi=(1)/(2pi)(m)/(s)`