Question

A ball is released from height 'h' which makes perfectly elastic collision with ground. The frequency of periodic vibratory motion is (g=acceleration due to gravity)

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

Hint:

One full oscillation is a fall and a rise, so the period is twice the fall time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A ball dropped from height \(h\) hits the ground and rebounds elastically to the same height. This repeats, so the motion is periodic.

Step 2: Time of one fall:
\[ h = \frac12 g t^2 \Rightarrow t = \sqrt{\frac{2h}{g}} \]

Step 3: Period and frequency:
One full cycle consists of a fall and a rise of equal time: \(T = 2\sqrt{\dfrac{2h}{g}}\).
\[ f = \frac1T = \frac12\sqrt{\frac{g}{2h}} \]

Step 4: Check the options:
Option (A) matches. Option (B) is \(\tfrac12\sqrt{2h/g}\), which is half the fall time and not a frequency. Options (C) and (D) contain \(2\pi\), which belongs to SHM and not to this motion.

Final Answer:
The period is twice the fall time. \[ \boxed{\text{(A) }\dfrac12\sqrt{\dfrac{g}{2h}}} \]