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

Explanation

Mean position of the particles is `(mg)/(K)` distance below unstretched position of spring. Therefore, amplitude of oscillation is
`A = (mg)/(K) • = sqrt((K)/(m)) = 2•f = 20•`
`:. (m)/(K) = (1)/(•^(2)), A = (g)/(•^(2))`
Therefore, the maximum speed of particle will be
`V_(max) = A• = (g)/(•^(2)) xx • =(g)/(•) = (1)/(2•)m//s`