Question

A particle of mass m performs SHM along a straight line with frequency f and amplitude A.

• A. The average kinetic energy of the particle is zero.
• B. The average potential energy is m ${\pi }^2{f}^2{A}^2$.
• C. The frequency of ocillation of kinetic energy is 2f.
• D. Velocity function leads acceleration by $\pi$/2.

Answer 1

Answer: (B), (C)

The correct options are
B The frequency of ocillation of kinetic energy is 2f.
C The average potential energy is m ${\pi }^2{f}^2{A}^2$.

$\text{Given:-}$ Mass of the particle = m

$\text{Solution:-}$ We know that,

$x=Asin\omega t$

For velocity,

v $=\frac{dx}{dt}=A\omega cos\omega t$

Squaring the above equation,

$v^2=A^2{\omega }^{2}co{s}^2\omega t$

$v^2=A^2(2\pi f{)}^{2}co{s}^2\omega t$ where $\omega =2\pi f$

$=4{\pi }^2{f}^2{A}^2\left(\frac{1+cos2\omega t}{2}\right)$

$=2{\pi }^2{f}^2{A}^2(1+cos2\omega t)$

Now, average potential energy = average kinetic energy

$\Rightarrow \frac{1}{2}m{v}^2=\frac{1}{2}m\left[2{\pi }^2{f}^2{A}^2(1+cos2\omega t)\right]=m{\pi }^2{f}^2{A}^2$

For angular frequency of kinetic energy $=2\omega$

So, frequency of oscillation of kinetic energy $=2f$

$\text{Hence the correct option is B and C}$