Question

The total energy of the particle executing SHM is directly proportional to the square of the following quantity.
A. Acceleration
B. Amplitude
C. Time period
D. Mass

Answer 1

Recall the velocity of the particle performing SHM and express the kinetic energy. Determine the total energy of the particle performing SHM by substituting the expressions for kinetic energy and potential energy. Verify the proportional dependence of the total energy using the formula you obtained.

Complete step by step answer:
To answer this question, we can derive the expression for total energy of the particle performing SHM.

Let the particle of mass m is performing SHM about the mean position. We assume at a given instant of time, the displacement of the particle is x from the mean position and the velocity of the particle is v.

We know the expression for the velocity of the particle performing SHM,
$v = \omega \sqrt {{A^2} - {x^2}}$ …… (1)

Here, $\omega$ is the angular frequency and A is the amplitude of oscillation.

We have the expression for the kinetic energy of the particle,
${E_K} = \dfrac{1}{2}m{v^2}$

Substituting the value of velocity from equation (1) in the above equation, we get,
${E_K} = \dfrac{1}{2}m{\left( {\omega \sqrt {{A^2} - {x^2}} } \right)^2}$
$\Rightarrow {E_K} = \dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right)$ …… (2)
We know that the potential energy of the particle in simple harmonic motion is given as,
${E_P} = \dfrac{1}{2}m{\omega ^2}{x^2}$ ……. (3)

The total energy of the particle is the sum of the kinetic energy and the potential energy of the particle. Therefore, we have from equation (2) and (3), the total energy of the particle,
$E = {E_K} + {E_P}$
$\Rightarrow E = \dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right) + \dfrac{1}{2}m{\omega ^2}{x^2}$
$\Rightarrow E = \dfrac{1}{2}m{\omega ^2}{A^2} - \dfrac{1}{2}m{\omega ^2}{x^2} + \dfrac{1}{2}m{\omega ^2}{x^2}$
$\Rightarrow E = \dfrac{1}{2}m{\omega ^2}{A^2}$ …… (4)

Therefore, from the above equation, the total energy of the particle performing SHM is proportional to the square of the amplitude of oscillations. The above equation does not have the term acceleration. Therefore, the option (A) is incorrect. From the above equation, the total energy of the particle is proportional to the mass and not the square of the mass. Therefore, the option (D) is incorrect.

We know that the angular frequency $\omega$ is expressed as,
$\omega = 2\pi n$

Therefore, substituting the above equation in equation (4), we get,
$E = \dfrac{1}{2}m{\left( {2\pi n} \right)^2}{A^2}$
$\Rightarrow E = \dfrac{{2{\pi ^2}m{A^2}}}{{{T^2}}}$, where T is the time period.

Therefore, the total energy is inversely proportional to the square of the time period. Thus, the option (C) is incorrect.

So, the correct answer is “Option B”.

Note:
We have the force acting on the particle performing simple harmonic motion is $F = kx$. Therefore, the acceleration of the particle can be expressed using Newton’s second law as, $a = \dfrac{{kx}}{m}$. Since the total energy of the particle is proportional to the force constant k, it is also proportional to the acceleration a of the particle.