Which of the following is true for total energy and potential energy of a particle in SHM, when the displacement is half the amplitude of oscillation?

The correct answer is option 1) i.e. The total energy of the particle is four times its potential energy.

CONCEPT:

• Simple harmonic motion (SHM): It is a type of oscillatory motion in which the restoring force is directly proportional to the displacement of the body from its mean position.​
• The energy in simple harmonic motion: 
  •  The total energy of a particle executing SHM is the sum of its kinetic and potential energy.

The kinetic energy is given by the equation: \(KE = \frac{1}{2}mω ^2 (a^2 - y^2)\)

The potential energy is given by the equation: \(PE = \frac{1}{2} mω ^2y^2\)

The total energy, E = KE + PE

\(E = \frac{1}{2}mω ^2 (a^2 - y^2) + \frac{1}{2} mω ^2y^2\)

\(⇒E = \frac{1}{2} mω ^2a^2\)

Where m is the mass, ω is the angular frequency, y is the displacement of the particle from the mean position and a is the amplitude.

EXPLANATION:

Given that:

Displacement is half the amplitude ⇒ y = a/2

\(PE = \frac{1}{2} mω ^2y^2 = \frac{1}{2} mω ^2(a/2)^2=\frac{1}{8} mω ^2a^2\)

\(E = \frac{1}{2} mω ^2a^2\)

Ratio \(= \frac{E}{PE} = \frac{\frac{1}{2} mω ^2a^2}{\frac{1}{8} mω ^2a^2} = \frac{4}{1}\)

Thus, the total energy is four times the potential energy.