Question

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

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

Hint:

The time period is the time to fall plus the time to rise back to the same height, which is twice the fall time.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
With a perfectly elastic collision, the ball returns to the same height and the motion repeats.

Step 2: Time of fall
From \(h=\frac{1}{2}gt^{2}\), the fall time is \(t=\sqrt{\frac{2h}{g}}\).

Step 3: Time period and frequency
The period is fall plus rise: \(T=2\sqrt{\frac{2h}{g}}\).
\(f=\frac{1}{T}=\frac{1}{2}\sqrt{\frac{g}{2h}}\)

Step 4: Compare the options
This is option (B).

Final Answer:
The frequency is one half times the square root of g over 2h. \[ \boxed{\text{(B) }\dfrac{1}{2}\sqrt{\dfrac{g}{2h}}} \]