Question

A particle is executing linear simple harmonic oscillation with an amplitude of \( A \). If the total energy of oscillation is \( E \), then its kinetic energy at a distance of \( 0.707A \) from the mean position is:

• A. \( \frac{E}{2} \)
• B. \( \frac{E}{4} \)
• C. \( \frac{3E}{4} \)
• D. \( \frac{E}{4} \)
• E. \( E \)

Hint:

In simple harmonic motion, the total energy is shared between kinetic and potential energies. The kinetic energy at any point can be found by subtracting potential energy from total energy.

Answer 1

The Correct Option is A

In simple harmonic motion (SHM), the total mechanical energy is conserved and is the sum of the kinetic energy \( K \) and potential energy \( U \).
The total energy \( E \) is constant and is given by: \[ E = K + U \] The total energy in SHM is also related to the amplitude \( A \) and is given by: \[ E = \frac{1}{2} m \omega^2 A^2 \] where \( m \) is the mass of the particle and \( \omega \) is the angular frequency. 
Step 1: Kinetic Energy in SHM The kinetic energy \( K \) of the particle at a position \( x \) from the mean position is given by: \[ K = \frac{1}{2} m v^2 \] where \( v \) is the velocity of the particle at position \( x \). The velocity in SHM is related to the displacement by: \[ v = \omega \sqrt{A^2 - x^2} \] Thus, the kinetic energy becomes: \[ K = \frac{1}{2} m \omega^2 \left( A^2 - x^2 \right) \] 
Step 2: Kinetic Energy at \( x = 0.707A \) At \( x = 0.707A \), the displacement is \( 0.707 \) times the amplitude. Substituting this into the expression for kinetic energy: \[ K = \frac{1}{2} m \omega^2 \left( A^2 - (0.707A)^2 \right) \] \[ K = \frac{1}{2} m \omega^2 \left( A^2 - 0.5A^2 \right) \] \[ K = \frac{1}{2} m \omega^2 \times 0.5A^2 \] \[ K = \frac{1}{4} m \omega^2 A^2 \] Since the total energy \( E = \frac{1}{2} m \omega^2 A^2 \), we have: \[ K = \frac{1}{2} E \] Thus, the kinetic energy at a distance of \( 0.707A \) from the mean position is \( \frac{E}{2} \).