Question

A particle executes simple harmonic motion with a frequency f. The frequency with which the kinetic energy oscillates is
[IIT JEE 1987]

• A. $\frac{f}{2}$
• B. f
• C. 2f
• D. 4f

Answer 1

The correct option is C

2f

So we have to find the frequency of K.E. Okay! Let's take the mean position as reference and try to figure out. When the particle crosses the mean position its K.E is maximum and P.E is zero so basically K.E. = T.E at this position. Lets say at t=0 the particle crosses the mean position its K.E. = K${}_{max}$. It would come back to mean position after time $\frac{T}{2}$ and have the same magnitude of velocity and hence same K.E. = K${}_{max}$ since K.E. "repeats” itself in half the time period of SHM its frequency will be twice i.e., 2f.