Question

A block of mass \( m \) attached to one end of the vertical spring produces extension \( x \). If the block is pulled and released, the periodic time of oscillation is

• A. \( 2\pi \sqrt{\frac{2x}{g}} \)
• B. \( 2\pi \sqrt{\frac{x}{g}} \)
• C. \( 2\pi \sqrt{\frac{x}{2g}} \)
• D. \( 2\pi \sqrt{\frac{x}{4g}} \)

Hint:

The period of oscillation for a mass-spring system depends on the spring constant and the mass, and is independent of the amplitude (for small oscillations).

Answer 1

The Correct Option is B

Step 1: Understanding the oscillatory motion.
The periodic time of oscillation of a block attached to a spring is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where \( m \) is the mass of the block and \( k \) is the spring constant.
Step 2: Relation between force and spring constant.
The force due to the spring is \( F = kx \), where \( x \) is the extension. Also, the weight of the block is \( mg \). So, at equilibrium: \[ kx = mg \implies k = \frac{mg}{x} \]
Step 3: Substituting the values.
Substitute the value of \( k \) into the formula for the time period: \[ T = 2\pi \sqrt{\frac{m}{\frac{mg}{x}}} = 2\pi \sqrt{\frac{x}{g}} \]
Step 4: Conclusion.
The periodic time of oscillation is \( 2\pi \sqrt{\frac{x}{g}} \), so the correct answer is (B).