Question

A body attached to a spring oscillates in horizontal plane with frequency \( n \). Its total energy is \( E \). If the velocity in the mean position is \( v \), then the spring constant is

• A. \( \frac{E\pi^2n^2}{v^2} \)
• B. \( \frac{4E\pi^2n^2}{v^2} \)
• C. \( \frac{8E\pi^2n^2}{v^2} \)
• D. \( \frac{2E\pi^2n^2}{v^2} \)

Hint:

In simple harmonic motion, the total energy is proportional to the square of the amplitude and the square of the frequency. The spring constant can be related to the energy and velocity.

Answer 1

The Correct Option is C

Step 1: Energy in simple harmonic motion (S.H.M.).
In simple harmonic motion, the total energy \( E \) is given by: \[ E = \frac{1}{2} m \omega^2 A^2 \] where \( m \) is the mass of the body, \( \omega \) is the angular frequency, and \( A \) is the amplitude of oscillation. Also, the velocity at the mean position is given by: \[ v = \omega A \] Step 2: Relating total energy and velocity.
We know that the angular frequency \( \omega = 2\pi n \), so the total energy can be rewritten as: \[ E = \frac{1}{2} m (2\pi n)^2 A^2 \] Substituting \( v = \omega A \), we get: \[ E = \frac{1}{2} m \left( \frac{v}{A} \right)^2 (2\pi n)^2 A^2 \] Step 3: Solving for spring constant.
The spring constant \( k \) is related to the angular frequency by: \[ k = m \omega^2 \] Substituting \( \omega = 2\pi n \), we get: \[ k = m (2\pi n)^2 \] Step 4: Conclusion.
Thus, the spring constant is \( \frac{8E\pi^2n^2}{v^2} \), corresponding to option (C).