The time period of a bob performing simple harmonic motion in water is 2s. If density of bob is `(4)/(3) xx 10^(3) kg// m^(3)` then time period of bob performing simple harmonic motion in air will be

Options

A. 3 s
B. 4 s
C. 2 s
D. 1 s

Explanation

To solve the problem, we need to find the time period of a bob performing simple harmonic motion in air, given its time period in water and its density.

### Step-by-Step Solution:

1. **Understanding the Time Period in Water:**
The time period \( T \) of a bob in simple harmonic motion is given by:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} \]
where \( g_{\text{eff}} \) is the effective acceleration due to gravity when the bob is submerged in water.

2. **Effective Gravity in Water:**
The effective gravity \( g_{\text{eff}} \) when the bob is in water can be expressed as:
\[ g_{\text{eff}} = g - \frac{\rho_{\text{liquid}}}{\rho_{\text{bob}}} g \]
where \( \rho_{\text{liquid}} \) is the density of water (approximately \( 10^3 \, \text{kg/m}^3 \)), and \( \rho_{\text{bob}} \) is the density of the bob.

3. **Given Values:**
- Time period in water \( T = 2 \, \text{s} \)
- Density of the bob \( \rho_{\text{bob}} = \frac{4}{3} \times 10^3 \, \text{kg/m}^3 \)
- Density of water \( \rho_{\text{liquid}} = 10^3 \, \text{kg/m}^3 \)

4. **Calculating Effective Gravity:**
Substitute the values into the effective gravity formula:
\[ g_{\text{eff}} = g - \frac{10^3}{\frac{4}{3} \times 10^3} g = g - \frac{3}{4} g = \frac{1}{4} g \]

5. **Substituting into the Time Period Equation:**
Now, substituting \( g_{\text{eff}} \) into the time period equation:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{L}{\frac{1}{4}g}} = 2\pi \sqrt{\frac{4L}{g}} = 2 \times 2\pi \sqrt{\frac{L}{g}} \]
This means:
\[ 2 = 4\pi \sqrt{\frac{L}{g}} \implies \sqrt{\frac{L}{g}} = \frac{1}{2\pi} \]

6. **Time Period in Air:**
The time period in air is given by:
\[ T_{\text{air}} = 2\pi \sqrt{\frac{L}{g}} \]
From our earlier calculation, we have \( \sqrt{\frac{L}{g}} = \frac{1}{2\pi} \):
\[ T_{\text{air}} = 2\pi \left(\frac{1}{2\pi}\right) = 1 \, \text{s} \]

### Final Answer:
The time period of the bob performing simple harmonic motion in air is \( \boxed{1 \, \text{s}} \).