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 of length \( \sqrt{5} \, \text{m} \) suspended in a car moving with uniform acceleration of \( 5 \, \text{m/s}^2 \), we can follow these steps:

### Step 1: Understand the effective gravitational acceleration
In a non-inertial frame (like a car accelerating horizontally), the effective gravitational acceleration \( g_{\text{effective}} \) can be calculated using the formula:
\[ g_{\text{effective}} = \sqrt{g^2 + a^2} \]
where:
- \( g \) is the acceleration due to gravity (given as \( 10 \, \text{m/s}^2 \)),
- \( a \) is the horizontal acceleration of the car (given as \( 5 \, \text{m/s}^2 \)).

### Step 2: Calculate \( g_{\text{effective}} \)
Substituting the values into the formula:
\[ g_{\text{effective}} = \sqrt{(10)^2 + (5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \, \text{m/s}^2 \]

### Step 3: Use the formula for the time period of a pendulum
The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
where \( L \) is the length of the pendulum.

### Step 4: Substitute the values into the time period formula
Here, \( L = \sqrt{5} \, \text{m} \) and \( g_{\text{effective}} = 5\sqrt{5} \, \text{m/s}^2 \):
\[ T = 2\pi \sqrt{\frac{\sqrt{5}}{5\sqrt{5}}} \]

### Step 5: Simplify the expression
\[ T = 2\pi \sqrt{\frac{1}{5}} = 2\pi \cdot \frac{1}{\sqrt{5}} = \frac{2\pi}{\sqrt{5}} \]

### Final Answer
Thus, the time period of the pendulum is:
\[ T = \frac{2\pi}{\sqrt{5}} \, \text{s} \]

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