A particle executing simple harmonic motion of amplirtude `5cm` has maximum speed of `31.4 cm//s` The frequency of its oscillation is

Options

A. `3Hz`
B. `2Hz`
C. `4Hz`
D. `1Hz`

Answer

D

Explanation

Maximum speed of a particle executing `SHM` is
`k_(max) = a omega = a(2pir)`
`rArr n = (v_(max))/(2pi a)`
Here `v_(max) = 31.4 cm//s a= 5cm`
subsititution the given values we have
`n = (31.4)/(2 xx 3.14 xx 5) = 1 Hz`