A pendulum suspended from the ceiling of train has a period T When the train is at rest. When the train si accelerating with a uniform acceleratio a, the period of oscillation will

Options

A. increase
B. decrease
C. remained unaffected
D. becomes infinite

Answer

To solve the problem, we need to analyze how the period of a pendulum changes when the train it is suspended from is accelerating.

### Step-by-Step Solution:

1. **Understanding the Period of a Pendulum at Rest**:
The period \( T_1 \) of a simple pendulum when the train is at rest is given by the formula:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]
where:
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

2. **Considering the Effect of Train's Acceleration**:
When the train accelerates with a uniform acceleration \( a \), the effective acceleration acting on the pendulum changes. The pendulum will experience a pseudo force due to the acceleration of the train.

3. **Calculating the New Effective Gravity**:
The effective acceleration due to gravity when the train is accelerating can be calculated as:
\[ g' = \sqrt{g^2 + a^2} \]
This is because the pendulum will experience both the gravitational force and the force due to the train's acceleration.

4. **Finding the New Period of the Pendulum**:
The new period \( T_2 \) of the pendulum when the train is accelerating is given by:
\[ T_2 = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \):
\[ T_2 = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + a^2}}} \]

5. **Comparing the Periods**:
Since \( g' = \sqrt{g^2 + a^2} \) is greater than \( g \) (because \( a \) is a positive value), it follows that:
\[ T_2 < T_1 \]
This indicates that the period of oscillation decreases when the train accelerates.

6. **Conclusion**:
Therefore, the period of oscillation of the pendulum when the train is accelerating is less than the period when the train is at rest.

### Final Answer:
The period of oscillation will **decrease**.

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