The time - period of a simple pendulum of length `sqrt5m` suspended in a car moving with uniform acceleration of `5ms^(-2)` in a horizontal straight road is
`(g=10ms^(-2))`

Options

A. `(2pi)/(sqrt5)s`
B. `(pi)/(sqrt5)s`
C. `5pis`
D. `4pis`

Answer

To find the time period of a simple pendulum suspended in a car that is accelerating, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the pendulum, \( L = \sqrt{5} \, \text{m} \)
- Acceleration of the car, \( a = 5 \, \text{m/s}^2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Determine the Effective Gravity:**
When the pendulum is in an accelerating car, we need to consider the effective gravity (\( g_{\text{effective}} \)). The effective gravity can be calculated using the formula:
\[ g_{\text{effective}} = \sqrt{g^2 + a^2} \]
Substituting the values:
\[ g_{\text{effective}} = \sqrt{(10)^2 + (5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \, \text{m/s}^2 \]

3. **Use the Formula for the Time Period of a Pendulum:**
The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
Substituting the values of \( L \) and \( g_{\text{effective}} \):
\[ T = 2\pi \sqrt{\frac{\sqrt{5}}{5\sqrt{5}}} \]

4. **Simplify the Expression:**
Simplifying the fraction inside the square root:
\[ T = 2\pi \sqrt{\frac{1}{5}} = 2\pi \cdot \frac{1}{\sqrt{5}} = \frac{2\pi}{\sqrt{5}} \, \text{seconds} \]

5. **Final Result:**
The time period of the pendulum is:
\[ T = \frac{2\pi}{\sqrt{5}} \, \text{seconds} \]