A body dropped from a height `h` onto the floow makes elastic collision with the floow. The frequency of oscillation of its periodic motion is

Options

A. `(1)/(2)sqrt((g)/(2h))`
B. `(1)/(2)sqrt((2h)/(g))`
C. `(1)/(2pi)sqrt((g)/(2h))`
D. `2pisqrt((g)/(2h))`

Explanation

To solve the problem of finding the frequency of oscillation of a body dropped from a height \( h \) that makes an elastic collision with the floor, we can follow these steps:

### Step 1: Understand the Motion
When the body is dropped from a height \( h \) and collides elastically with the floor, it will fall under the influence of gravity and then bounce back to the same height due to the elastic nature of the collision. The motion can be modeled as a periodic motion.

### Step 2: Determine the Time Period
The time period \( T \) of the oscillation is the total time taken for the body to fall to the ground and then return to its original height. This can be divided into two parts:
- Time taken to fall from height \( h \) to the ground.
- Time taken to rise back from the ground to height \( h \).

### Step 3: Calculate the Time to Fall
Using the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
where:
- \( S = h \) (the distance fallen),
- \( u = 0 \) (initial velocity, since it is dropped),
- \( a = g \) (acceleration due to gravity).

Substituting these values, we get:
\[ h = 0 \cdot t + \frac{1}{2} g t^2 \]
This simplifies to:
\[ h = \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t^2 = \frac{2h}{g} \]
Taking the square root:
\[ t = \sqrt{\frac{2h}{g}} \]

### Step 4: Calculate the Total Time Period
Since the time to fall is \( t \) and the time to rise back is also \( t \), the total time period \( T \) is:
\[ T = t + t = 2t = 2\sqrt{\frac{2h}{g}} \]

### Step 5: Calculate the Frequency
The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{1}{2\sqrt{\frac{2h}{g}}} \]
This can be simplified to:
\[ f = \frac{1}{2} \sqrt{\frac{g}{2h}} \]

### Final Answer
Thus, the frequency of oscillation of the body is:
\[ f = \frac{1}{2} \sqrt{\frac{g}{2h}} \]

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