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, we need to determine the angle \(\theta\) at which the bob of mass \(m\) will oscillate when the car is moving with a constant horizontal acceleration \(a\). Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Bob
When the car accelerates horizontally, the bob experiences two forces:
1. The gravitational force acting downwards, which is \(mg\).
2. A pseudo force acting horizontally in the opposite direction of the car's acceleration, which is \(ma\).

### Step 2: Draw a Free Body Diagram
Visualize the situation:
- The bob is suspended from the ceiling of the car.
- The gravitational force acts vertically downward.
- The pseudo force due to the car's acceleration acts horizontally.

### Step 3: Establish the Relationship between Forces
The bob will hang at an angle \(\theta\) from the vertical. The forces can be represented as:
- The vertical component of the tension in the string balances the weight of the bob: \(T \cos(\theta) = mg\).
- The horizontal component of the tension balances the pseudo force: \(T \sin(\theta) = ma\).

### Step 4: Divide the Two Equations
To eliminate \(T\) (the tension in the string), we can divide the two equations:
\[ \frac{T \sin(\theta)}{T \cos(\theta)} = \frac{ma}{mg} \]
This simplifies to:
\[ \tan(\theta) = \frac{a}{g} \]

### Step 5: Solve for \(\theta\)
To find the angle \(\theta\), we take the arctangent of both sides:
\[ \theta = \tan^{-1}\left(\frac{a}{g}\right) \]

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