The period of oscillation of a simple pendulum of length `L` suspended from the roof of a vehicle which moves without frication down on inclined plane of inclination `alpha = 60^(@)` is given by `pisqrt((XL)/(g))` then find `X`.

Explanation

To find the value of \( X \) in the period of oscillation of a simple pendulum suspended from a vehicle moving down an inclined plane, we can follow these steps:

### Step 1: Understand the Forces Acting on the Pendulum
The pendulum experiences gravitational force \( mg \) acting downwards. When the vehicle is on an inclined plane with an angle \( \alpha = 60^\circ \), we can resolve the gravitational force into two components:
- Perpendicular to the incline: \( mg \cos(60^\circ) \)
- Parallel to the incline: \( mg \sin(60^\circ) \)

### Step 2: Determine the Effective Gravity
The tension \( T \) in the pendulum string is affected by the effective gravitational force acting on it. The effective gravitational acceleration \( g' \) for the pendulum can be determined as follows:
- The tension \( T \) in the pendulum is equal to the component of gravitational force acting perpendicular to the incline, which is \( mg \cos(60^\circ) \).
- Since \( \cos(60^\circ) = \frac{1}{2} \), we have:
\[ T = mg \cos(60^\circ) = mg \cdot \frac{1}{2} = \frac{mg}{2} \]

### Step 3: Relate Tension to Effective Gravity
The effective gravitational acceleration \( g' \) acting on the pendulum is given by:
\[ g' = \frac{T}{m} = \frac{\frac{mg}{2}}{m} = \frac{g}{2} \]

### Step 4: Write the Formula for the Period of the Pendulum
The period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' = \frac{g}{2} \):
\[ T = 2\pi \sqrt{\frac{L}{\frac{g}{2}}} = 2\pi \sqrt{\frac{2L}{g}} \]

### Step 5: Compare with the Given Period Formula
We are given that the period can also be expressed as:
\[ T = \pi \sqrt{\frac{X L}{g}} \]
By comparing the two expressions for the period:
\[ 2\pi \sqrt{\frac{2L}{g}} = \pi \sqrt{\frac{X L}{g}} \]

### Step 6: Simplify and Solve for \( X \)
Dividing both sides by \( \pi \) and \( \sqrt{L/g} \):
\[ 2\sqrt{2} = \sqrt{X} \]
Squaring both sides gives:
\[ 4 \cdot 2 = X \quad \Rightarrow \quad X = 8 \]

### Final Answer
Thus, the value of \( X \) is \( 8 \).

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