Question

Frequency of variation of kinetic energy of a simple harmonic motion of frequency n is

• A. $2$n
• B. $n$
• C. $\frac{n}{2}$
• D. $3$n

Answer 1

Answer: (A)

The velocity of a simple harmonic oscillator varies as $v_{0}sin(2\pi nt+\varphi )$

Kinetic energy is,

$K=\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_0^{2}si{n}^2(2\pi nt+\varphi )$

Now,

$si{n}^2\theta =\frac{1}{2}(1-cos(2\theta \left)\right)$

So,

$K=\frac{1}{2}m{v}_0^2(\frac{1}{2}-\frac{1}{2}cos(4\pi nt+2\varphi \left)\right)$

$K=\frac{1}{4}m{v}_0^2-\frac{1}{4}m{v}_0^{2}cos\left(2\pi \right(2n)t+2\varphi )$

Here, the time dependence comes from the last term and its frequency is $2n$.