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: Use the velocity at the ground:
Speed at impact $v = \sqrt{2gh}$. After the elastic bounce it leaves at the same speed.

Step 2: Time up and down:
Time to rise back to $h$ equals $v/g$, and the fall also takes $v/g$, so $T = 2v/g = 2\sqrt{2gh}/g = 2\sqrt{2h/g}$.

Step 3: Frequency:
$f = 1/T = \dfrac12\sqrt{\dfrac{g}{2h}}$, option (A).

Final Answer:
The frequency is (1/2) times root of g over 2h. \[ \boxed{\text{(A) }\dfrac12\sqrt{\dfrac{g}{2h}}} \]