A simple pendulum with a bob of mass `m` is suspended from the roof of a car moving with horizontal acceleration a

Options

A. The string makes an angle of `tan^(-1)(a//g)` with the vertical
B. The string makes an angle of `sin^(-1)((a)/(g))` with the vertical
C. The tension in the string is `msqrt(a^(2) + g^(2))`
D. The tension in the string is `msqrt(g^(2) - a^(2))`

Explanation

To solve the problem of a simple pendulum with a bob of mass `m` suspended from the roof of a car moving with horizontal acceleration `a`, we can follow these steps:

### Step 1: Understand the Forces Acting on the Bob
When the car accelerates horizontally, the bob will experience a pseudo force acting in the opposite direction of the car's acceleration. The forces acting on the bob are:
- The gravitational force (weight) acting downwards: \( mg \)
- The tension in the string acting at an angle \( \theta \) to the vertical.

### Step 2: Draw the Free Body Diagram
In the free body diagram, we can resolve the tension \( T \) into two components:
- Vertical component: \( T \cos \theta \)
- Horizontal component: \( T \sin \theta \)

### Step 3: Set Up the Equations
From the free body diagram, we can write two equations based on the balance of forces:
1. In the vertical direction (for equilibrium):
\[ T \cos \theta = mg \quad (1) \]
2. In the horizontal direction (due to pseudo force):
\[ T \sin \theta = ma \quad (2) \]

### Step 4: Divide the Equations
To eliminate \( T \) and find the relationship between \( a \) and \( g \), we can 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: Find the Angle \( \theta \)
From the above equation, we can express \( \theta \) as:
\[ \theta = \tan^{-1}\left(\frac{a}{g}\right) \]

### Step 6: Find the Tension \( T \)
Now we can substitute \( \theta \) back into one of our original equations to find the tension \( T \). Using equation (1):
\[ T = \frac{mg}{\cos \theta} \]
Using the identity \( \cos \theta = \frac{g}{\sqrt{a^2 + g^2}} \) (derived from the triangle formed by \( a \) and \( g \)):
\[ T = mg \cdot \frac{\sqrt{a^2 + g^2}}{g} = m \sqrt{a^2 + g^2} \]

### Final Result
Thus, the tension in the string is:
\[ T = m \sqrt{a^2 + g^2} \]