Frequency of kinetic energy in shm I am currently learning about simple harmonic motion.
In a book I am reading it says frequency of kinetic energy is twice the frequency of velocity for a harmonic oscillator by showing velocity vs time graph and KE vs time graph
What I think it means to say is that kinetic energy reaches its maximum value 2 times in the same time it takes for velocity to reach its maximum value.
Am I right?
 A: Yes, you are right in a time period $T$ let's say a particle moves from one extreme to another and back of a simple pendulum. During this time it achieves maximum velocity say $v$ two times but it is in opposite directions. Kinetic energy however does not depend on direction of velocity as it depends on $v^2$, hence in the same time period it is achieved 2 times, hence its frequency is twice that of velocity.
A: In SHM the velocity is given by
$v(t) = \dfrac{\mathrm{d} x}{\mathrm{d} t} = - A\omega \sin(\omega t+\varphi),$
The time period of $v(t)$ is nothing but the time period of $sin(\omega t)$ function which is $2\pi/\omega$.
Now the $K.E$ is $\dfrac{1}{2} m v^2$ which is given by 
 $K(t) = \frac{1}{2} mv^2(t) = \frac{1}{2}m\omega^2A^2\sin^2(\omega t + \varphi) )$
Now the time period of $K.E$ is nothing but the time period of $sin^2(\omega t)$ which is $\pi/\omega$.
Frequency of any function is inverse of time period so
frequency of $K.E$ = $\dfrac{\omega}{\pi}$
and frequency of velocity $v(t)$ = $\dfrac{\omega}{2\pi}$
,that is frequency of kinetic energy is double than that of the velocity.
