Question

A metal ball released from height \( h \) makes perfectly elastic collision with ground. The frequency of periodic vibratory motion is

• A. \( \frac{1}{2\pi} \sqrt{\frac{g}{2h}} \)
• B. \( \frac{1}{2} \sqrt{\frac{g}{2h}} \)
• C. \( \frac{1}{2} \sqrt{\frac{2h}{g}} \)
• D. \( \frac{1}{2\pi} \sqrt{\frac{2h}{g}} \)

Hint:

The frequency of periodic motion due to free fall depends on the height from which the object is released.

Answer 1

The Correct Option is B

Step 1: Understanding the periodic vibratory motion.
When the ball hits the ground and rebounds, it undergoes periodic vibratory motion. The frequency \( f \) of this motion is related to the time period \( T \) by: \[ f = \frac{1}{T} \] where the time period is related to the height from which the ball falls.
Step 2: Finding the time period.
For a free fall, the time taken to fall from height \( h \) is given by: \[ T = \sqrt{\frac{2h}{g}} \] Step 3: Conclusion.
The frequency is the reciprocal of the time period: \[ f = \frac{1}{2\pi} \sqrt{\frac{g}{2h}} \] Thus, the correct answer is (B).