The position of a particle axecuting simple harmonic motion is given by
`x (t) = 2 cos ((pi)/(15) t - (pi)/(2)), ` where x is in centimetre and t is in seconds. The time period of the kinetic energy of the particle in second is

Options

A. `pi`
B. `pi/15`
C. `15`
D. 30

Answer

C

Explanation

Given, position in `SHM, x = 2 cos ((pi)/(15)t - (pi)/(2))`
On comparing it with, we get
`x = A cos (omega t - (pi)/(2))`
Angular frequency, `omega = (pi)/(15)`
` (2pi)/(T) = (pi)/(15) implies T =30`
So, time period of SHM is `T =30s,` but time period of KE would be `T/2, i.e. 15 s.`