The time period of a simple pendulum in a stationary train is T. The time period of a mass attached to a spring is also T. The train accelerates at the rate `5 m//s^(2)`. If the new time periods of the pendulum and spring be `T_(P)` and `T_(S)` respectively, then :-

Options

A. `T_(P)=T_(S)`
B. `T_(P)gtT_(S)`
C. `T_(P)ltT_(S)`
D. Cannot be predicted

Answer

To solve the problem, we need to analyze how the time periods of a simple pendulum and a mass attached to a spring change when the train accelerates.

### Step 1: Understand the Time Period of a Simple Pendulum
The time period \( T \) of a simple pendulum 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.

### Step 2: Determine the Effect of Train Acceleration on the Pendulum
When the train accelerates at \( a = 5 \, \text{m/s}^2 \), the effective acceleration due to gravity \( g' \) experienced by the pendulum changes. The effective gravitational acceleration can be calculated as:
\[ g' = g + a \]
This means that the effective gravitational force acting on the pendulum increases due to the acceleration of the train.

### Step 3: Calculate the New Time Period of the Pendulum
The new time period \( T_P \) of the pendulum when the train is accelerating can be expressed as:
\[ T_P = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \) into the equation gives:
\[ T_P = 2\pi \sqrt{\frac{L}{g + a}} \]
Since \( a = 5 \, \text{m/s}^2 \), we can see that \( g' > g \), which implies:
\[ T_P < T \]
Thus, the time period of the pendulum decreases when the train accelerates.

### Step 4: Understand the Time Period of a Mass on a Spring
The time period \( T_S \) of a mass attached to a spring is given by:
\[ T_S = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass and \( k \) is the spring constant. Importantly, the time period of a mass on a spring is independent of the acceleration due to gravity.

### Step 5: Determine the New Time Period of the Spring
Since the acceleration of the train does not affect the spring's time period, we have:
\[ T_S = T \]

### Step 6: Compare the Time Periods
From the above analysis, we conclude:
\[ T_P < T \quad \text{and} \quad T_S = T \]
Thus, we can state that:
\[ T_P < T_S \]

### Final Answer
The new time periods of the pendulum and spring are such that:
\[ T_P < T_S \]