If a ball is dropped from height 2 metre on a smooth eleastic floor, then the time period of oscillation is

Options

A. `(2)/(g )`
B. `(2)/( sqrt(g))`
C. `(4)/(g )`
D. `(4)/(sqrt(g))`

Answer

To find the time period of oscillation for a ball dropped from a height of 2 meters onto a smooth elastic floor, we can follow these steps:

### Step 1: Understand the Problem
The ball is dropped from a height of 2 meters onto an elastic floor. Since the floor is elastic, we can assume that there is no loss of energy during the collision. This means that the ball will bounce back to the same height from which it was dropped.

### Step 2: Calculate the Time Taken to Fall
The time taken for the ball to fall to the ground can be calculated using the formula for free fall:
\[ t_{\text{down}} = \sqrt{\frac{2h}{g}} \]
where:
- \( h = 2 \) meters (the height from which the ball is dropped)
- \( g \approx 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ t_{\text{down}} = \sqrt{\frac{2 \times 2}{9.81}} = \sqrt{\frac{4}{9.81}} \approx 0.64 \, \text{s} \]

### Step 3: Calculate the Time Taken to Rise
Since the floor is elastic, the time taken for the ball to rise back to the original height will be the same as the time taken to fall:
\[ t_{\text{up}} = t_{\text{down}} \approx 0.64 \, \text{s} \]

### Step 4: Calculate the Total Time for One Complete Oscillation
The total time for one complete oscillation (down and back up) is given by:
\[ T = t_{\text{down}} + t_{\text{up}} = t_{\text{down}} + t_{\text{down}} = 2t_{\text{down}} \]
Substituting the value of \( t_{\text{down}} \):
\[ T = 2 \times 0.64 \approx 1.28 \, \text{s} \]

### Step 5: Final Result
Thus, the time period of oscillation is approximately:
\[ T \approx 1.28 \, \text{s} \]