A simple pendulum is suspended from the ceiling of a car and its period of oscillation is T when the car is at rest. The car starts moving on a horizontal road with a constant acceleration g (equal to the acceleration due to gravity, in magnitude) in the forward direction. To keep the time period same, the length of th pendulum

Options

A. will have to be increased by `sqrt(2)l`
B. will have to be increased by `(sqrt(2)-1)l`
C. will have to be decreased by `sqrt(2)l`
D. will have to be decreased by `(sqrt(2)-1)l`

Explanation

To solve the problem, we need to analyze the situation of a simple pendulum in a car that is accelerating horizontally with an acceleration equal to the acceleration due to gravity (g). We will derive the new length of the pendulum required to keep the time period the same when the car is in motion.

### Step-by-Step Solution:

1. **Understanding the Time Period of a Simple Pendulum at Rest:**
The time period \( T \) of a simple pendulum at rest is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Analyzing the Situation When the Car is Accelerating:**
When the car accelerates with an acceleration \( g \), the pendulum experiences two forces: the gravitational force \( mg \) acting downwards and a pseudo force \( ma \) acting horizontally backward due to the car's acceleration. Here, \( a = g \).

3. **Finding the Effective Gravity:**
The effective gravitational force acting on the pendulum can be found using the Pythagorean theorem since the forces are perpendicular:
\[ g_{\text{effective}} = \sqrt{g^2 + g^2} = \sqrt{2g^2} = g\sqrt{2} \]

4. **New Time Period of the Pendulum in the Accelerating Car:**
The new time period \( T' \) of the pendulum when the car is accelerating is given by:
\[ T' = 2\pi \sqrt{\frac{L'}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{L'}{g\sqrt{2}}} \]

5. **Setting the Time Periods Equal:**
To keep the time period the same, we set \( T = T' \):
\[ 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{L'}{g\sqrt{2}}} \]
Canceling \( 2\pi \) from both sides gives:
\[ \sqrt{\frac{L}{g}} = \sqrt{\frac{L'}{g\sqrt{2}}} \]

6. **Simplifying the Equation:**
Squaring both sides results in:
\[ \frac{L}{g} = \frac{L'}{g\sqrt{2}} \]
Multiplying both sides by \( g\sqrt{2} \) gives:
\[ L\sqrt{2} = L' \]

7. **Final Result:**
Therefore, to keep the time period the same, the length of the pendulum must be increased to:
\[ L' = L\sqrt{2} \]

### Conclusion:
The length of the pendulum must be increased to \( L\sqrt{2} \) to maintain the same period of oscillation when the car accelerates with an acceleration equal to \( g \).

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