Question

A block of mass \( M \) hangs from a spring and oscillates vertically with an angular frequency \( \omega \). If the block is removed from the spring, when it is in equilibrium position, the spring shortens by:

• A. \( \frac{g}{\omega} \)
• B. \( \sqrt{\frac{g}{\omega}} \)
• C. \( \frac{g}{\omega^2} \)
• D. \( \frac{g}{\omega^2} \)

Hint:

The extension (or compression) of the spring in equilibrium is related to the gravitational force and the spring constant. Use the relationship \( k = M \omega^2 \) to solve for the displacement.

Answer 1

The Correct Option is C

Step 1: Understand the forces at equilibrium.
At equilibrium, the spring force balances the gravitational force: \[ k \Delta x = Mg \] where \( k \) is the spring constant and \( \Delta x \) is the spring compression (or extension).
Step 2: Relate the spring constant to angular frequency.
For SHM, the angular frequency is related to the spring constant by: \[ \omega = \sqrt{\frac{k}{M}} \quad \Rightarrow \quad k = M \omega^2 \]
Step 3: Solve for the compression \( \Delta x \). Substitute \( k = M \omega^2 \) into the equilibrium equation: \[ M \omega^2 \Delta x = Mg \] Solving for \( \Delta x \): \[ \Delta x = \frac{g}{\omega^2} \]