Question

The variations of kinetic energy \( K \), potential energy \( U \), and total energy \( E \) as a function of displacement of a particle in SHM is shown in the figure. The value of \( x_3 \) is 

• A. \( \frac{A}{2} \)
• B. \( \frac{A}{\sqrt{2}} \)
• C. \( A \)
• D. \( \frac{A}{3} \)

Hint:

In SHM, the kinetic and potential energies are equal when the displacement is \( \frac{A}{\sqrt{2}} \), where \( A \) is the amplitude.

Answer 1

The Correct Option is B

In simple harmonic motion (SHM), the total mechanical energy \( E \) is the sum of the kinetic energy \( K \) and the potential energy \( U \): \[ E = K + U \] For SHM, the total energy is constant and is given by: \[ E = \frac{1}{2} m \omega^2 A^2 \] where \( A \) is the amplitude of oscillation. The kinetic energy and potential energy vary as functions of the displacement \( x \). - At the maximum displacement (when \( x = \pm A \)), the potential energy is maximum, and the kinetic energy is zero. - At the equilibrium position (when \( x = 0 \)), the kinetic energy is maximum, and the potential energy is zero. The point where the kinetic energy is equal to the potential energy occurs when: \[ K = U \] Since \( K + U = E \), at this point: \[ K = U = \frac{E}{2} \] This condition occurs at \( x = \frac{A}{\sqrt{2}} \), which is the displacement where the kinetic and potential energies are equal. Thus, the correct value of \( x_3 \) is: \[ \frac{A}{\sqrt{2}} \]