Question

The kinetic energy of a particle in SHM is 8 J at its mean position. If its mass is 4 kg and amplitude is 1 m, then what is its time period?
(A) $\pi$ seconds
(B) $2\pi$ seconds
(C) $\dfrac{\pi }{2}$ seconds
(D) $4\pi$ seconds

Answer 1

Hint
The kinetic energy in a simple harmonic motion is dependent both on the angular frequency of oscillations and the amplitude of the wave. Since the angular frequency is given in radians, it is related to the time period inversely.
$\Rightarrow K = \dfrac{1}{2}m{\omega ^2}({A^2} - {y^2})$, where K is the kinetic energy in simple harmonic motion, m is the mass of the particle, A is the amplitude, y is the distance from the mean position and $\omega$ is the angular frequency.

Complete step by step answer
We are required to find the time period of oscillation of a particle exhibiting simple harmonic motion. We know that the kinetic energy for this kind of motion is given as:
$\Rightarrow K = \dfrac{1}{2}m{\omega ^2}({A^2} - {y^2})$
According to the question, we have the following information:
Kinetic energy $K = 8$ J
Mass of the particle $m = 4$ kg
Amplitude $A = 1$ m
Since we know that the particle is at mean position, y will be 0.
Now, putting these values in our main equation, we get:
$\Rightarrow 8 = \dfrac{1}{2}4{\omega ^2}({1^2} - {0^2}) = 2{\omega ^2}$
This gives us:
$\Rightarrow {\omega ^2} = 4$
$\Rightarrow \omega = 2$ [Since, the angular frequency cannot be negative]
We know that the angular frequency can be converted to the linear time period using the relation:
$\Rightarrow T = \dfrac{{2\pi }}{\omega } = \dfrac{{2\pi }}{2} = \pi$
This gives us the correct answer as option A.

Note
The simple harmonic motion is a special type of periodic motion. Here the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position. The oscillatory motion possessed by a pendulum or a person trying out bungee jumping is an example of simple harmonic motion.