A car is moving on a horizontal road with constant acceleration 'a' . A bob of mass 'm' is suspended from the ceiling of car. The mean position about which the bob will oscillate is given by `( 'theta'` is angle with vertical)

Options

A. `tan theta= (g)/(a)`
B. ` tantheta= (a)/(g)`
C. `tan theta = (2a)/9g`
D. `tan theta= (a)/(2g)`

Answer

To solve the problem of finding the mean position about which the bob will oscillate in a car moving with constant acceleration \( a \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Bob
When the car accelerates, the bob will not hang vertically downwards but will instead make an angle \( \theta \) with the vertical. The forces acting on the bob are:
- The tension \( T \) in the string acting along the string.
- The weight \( mg \) acting vertically downwards.

### Step 2: Draw the Free Body Diagram
Draw a free body diagram of the bob. The tension \( T \) can be resolved into two components:
- A vertical component \( T \cos \theta \) that balances the weight \( mg \).
- A horizontal component \( T \sin \theta \) that provides the necessary force for the horizontal acceleration of the bob.

### Step 3: Set Up the Equations of Motion
From the equilibrium condition in the vertical direction:
\[ T \cos \theta = mg \quad \text{(1)} \]
From Newton's second law in the horizontal direction:
\[ T \sin \theta = ma \quad \text{(2)} \]

### Step 4: Divide the Two Equations
To eliminate \( T \), divide equation (2) by equation (1):
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{ma}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{a}{g} \]

### Step 5: Conclusion
Thus, the angle \( \theta \) that the bob makes with the vertical is given by:
\[ \tan \theta = \frac{a}{g} \]

### Final Answer
The mean position about which the bob will oscillate is given by:
\[ \theta = \tan^{-1}\left(\frac{a}{g}\right) \]

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