Physics - SHM Question with Solution | TestHub

The displacement of a simple pendulum executing simple harmonic motion is given by $x=A\sin (\omega t+\varphi )$.

The velocity of the pendulum is $v=\frac{dx}{dt}=A\omega \cos (\omega t+\varphi )$.

The kinetic energy is given by $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t+\varphi )$.

Since kinetic energy is proportional to the square of velocity, it can never be negative. This eliminates the graphs that show negative values.

Also, the kinetic energy of a pendulum is not constant, which eliminates the constant graph.

Using the trigonometric identity ${\cos }^2\theta =\frac{1+\cos (2\theta )}{2}$, we get:

$K=\frac{1}{4}m{A}^2{\omega }^2[1+\cos (2\omega t+2\varphi )]$

This shows that the kinetic energy oscillates with an angular frequency $2\omega$, meaning its time period is half that of the pendulum, i.e., $T^{\prime }=\frac{T}{2}$.

The graph in option (3) correctly represents a non-negative oscillating quantity with a period of $T/2$.

Answer: