A pendulum suspended from the ceiling of the train has a time period of two seconds when the train is at rest, then the time period of the pendulum, if the train accelerates 10 `m//s^(2)` will be `(g=10m//s^(2))`

Options

A. 2s
B. `2 sqrt(2) s`
C. `2//sqrt(2) s`
D. `2^(3//4)`

Explanation

To solve the problem, we need to determine the new time period of a pendulum when the train it is suspended from accelerates. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the time period of a pendulum at rest
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: Use the given time period when the train is at rest
We know that when the train is at rest, the time period \( T_1 \) is 2 seconds. Therefore, we can write:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]
Given \( T_1 = 2 \) seconds and \( g = 10 \, \text{m/s}^2 \), we can set up the equation:
\[ 2 = 2\pi \sqrt{\frac{L}{10}} \]

### Step 3: Solve for \( L \)
To find \( L \), we can rearrange the equation:
\[ \sqrt{\frac{L}{10}} = \frac{2}{2\pi} \]
Squaring both sides gives:
\[ \frac{L}{10} = \left(\frac{1}{\pi}\right)^2 \]
Thus,
\[ L = 10 \left(\frac{1}{\pi}\right)^2 \]

### Step 4: Determine the effective acceleration due to gravity when the train accelerates
When the train accelerates at \( a = 10 \, \text{m/s}^2 \), the effective gravitational acceleration \( g_{\text{effective}} \) can be calculated using the Pythagorean theorem, since the gravitational force and the train's acceleration are perpendicular to each other:
\[ g_{\text{effective}} = \sqrt{g^2 + a^2} \]
Substituting the values:
\[ g_{\text{effective}} = \sqrt{10^2 + 10^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{m/s}^2 \]

### Step 5: Calculate the new time period when the train is accelerating
Now we can find the new time period \( T_2 \) using the effective gravitational acceleration:
\[ T_2 = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
Substituting \( g_{\text{effective}} = 10\sqrt{2} \):
\[ T_2 = 2\pi \sqrt{\frac{L}{10\sqrt{2}}} \]
Using \( L = 10 \left(\frac{1}{\pi}\right)^2 \):
\[ T_2 = 2\pi \sqrt{\frac{10 \left(\frac{1}{\pi}\right)^2}{10\sqrt{2}}} = 2\pi \sqrt{\frac{1}{\pi^2 \sqrt{2}}} \]
This simplifies to:
\[ T_2 = 2\pi \cdot \frac{1}{\pi} \cdot \frac{1}{\sqrt[4]{2}} = \frac{2}{\sqrt[4]{2}} = 2 \cdot 2^{-\frac{1}{4}} = 2^{1 - \frac{1}{4}} = 2^{\frac{3}{4}} \, \text{seconds} \]

### Final Answer
The new time period of the pendulum when the train accelerates at \( 10 \, \text{m/s}^2 \) is:
\[ T_2 = 2^{\frac{3}{4}} \, \text{seconds} \]