From the ceiling of a train, a pendulum of length 'l' is suspended. The train is moving with an acceleration `a_(0)` on horizontal surface. What must be the period of oscillation of pendulum?

Options

A. `T=2pisqrt(((l)/(g)))`
B. `T=2pisqrt(((l)/(a_(0)^(2)-g^(2))))`
C. `T=pisqrt(((l)/(a_(0)^(2)+g^(2))))`
D. `T=2pisqrt(((l)/(a_(0)^(2)-g^(2))))`

Explanation

To find the period of oscillation of a pendulum suspended from the ceiling of a train that is accelerating horizontally, we can follow these steps:

### Step 1: Understand the Effective Acceleration
When the train accelerates with an acceleration \( a_0 \), the pendulum experiences two accelerations:
1. The gravitational acceleration \( g \) acting downwards.
2. The train's acceleration \( a_0 \) acting horizontally.

### Step 2: Determine the Resultant Acceleration
The effective acceleration \( g_{\text{effective}} \) that affects the pendulum's motion can be determined using vector addition. The resultant acceleration can be calculated using the Pythagorean theorem:

\[ g_{\text{effective}} = \sqrt{a_0^2 + g^2} \]

### Step 3: Use the Formula for the Period of a Pendulum
The formula for the period \( T \) of a simple pendulum is given by:

\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]

### Step 4: Substitute the Effective Acceleration
Now we can substitute the expression for \( g_{\text{effective}} \) into the period formula:

\[ T = 2\pi \sqrt{\frac{L}{\sqrt{a_0^2 + g^2}}} \]

### Step 5: Simplify the Expression
We can simplify the expression further:

\[ T = 2\pi \frac{\sqrt{L}}{\sqrt{a_0^2 + g^2}} \]

### Final Result
Thus, the period of oscillation of the pendulum is:

\[ T = 2\pi \frac{\sqrt{L}}{\sqrt{a_0^2 + g^2}} \]