A pendulum is hanging from the ceiling of a cage. When the cage is moving up with certain acceleration and when it is moving down with the same acceleration, the tensions in the string are `T_(1)` and `T_2` respectively. When the cage moves horizontally with the same acceleration, the tension in the string is,

Options

A. `sqrt(2(T_(1)^(2) + T_(2)^(2))`
B. `sqrt((T_(1)^(2) + T_(2)^(2))/2)`
C. `sqrt(2(T_(1)^(2) -T_(2)^(2))`
D. `sqrt(T_(1)^(2) + T_(2)^(2))`

Answer

To solve the problem, we need to analyze the forces acting on the pendulum bob when the cage is moving in different directions. Let's break it down step by step.

### Step 1: Analyze the forces when the cage is moving upwards.
When the cage is moving upwards with acceleration \( A \), the forces acting on the pendulum bob are:
- The gravitational force \( mg \) acting downwards.
- The tension \( T_1 \) acting upwards.

Using Newton's second law, we can write the equation of motion:
\[ T_1 - mg = -ma \]
This can be rearranged to find \( T_1 \):
\[ T_1 = mg + ma = m(g + A) \]

### Step 2: Analyze the forces when the cage is moving downwards.
When the cage is moving downwards with the same acceleration \( A \), the forces acting on the pendulum bob are:
- The gravitational force \( mg \) acting downwards.
- The tension \( T_2 \) acting upwards.

Using Newton's second law again, we can write the equation of motion:
\[ T_2 - mg = ma \]
This can be rearranged to find \( T_2 \):
\[ T_2 = mg - ma = m(g - A) \]

### Step 3: Analyze the forces when the cage is moving horizontally.
When the cage is moving horizontally with acceleration \( A \), the pendulum will still experience gravitational force downwards and tension in the string. However, the tension will have both vertical and horizontal components due to the angle \( \theta \) that the string makes with the vertical.

The vertical component of tension must balance the weight:
\[ T \cos \theta = mg \]
The horizontal component of tension provides the necessary centripetal force:
\[ T \sin \theta = ma \]

### Step 4: Relate the components of tension.
From the two equations, we can express \( T \) in terms of \( m \), \( g \), and \( A \):
1. From \( T \cos \theta = mg \):
\[ T = \frac{mg}{\cos \theta} \]
2. From \( T \sin \theta = ma \):
\[ T = \frac{ma}{\sin \theta} \]

### Step 5: Combine the equations.
Since both expressions equal \( T \), we can set them equal to each other:
\[ \frac{mg}{\cos \theta} = \frac{ma}{\sin \theta} \]
Cross-multiplying gives:
\[ mg \sin \theta = ma \cos \theta \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ g \sin \theta = a \cos \theta \]

### Step 6: Find the tension when the cage moves horizontally.
To find the tension \( T \) in terms of \( g \) and \( A \), we can use the Pythagorean theorem:
\[ T^2 = (mg)^2 + (ma)^2 \]
Taking the square root gives:
\[ T = m \sqrt{g^2 + A^2} \]

### Final Answer:
Thus, the tension in the string when the cage is moving horizontally with acceleration \( A \) is:
\[ T = m \sqrt{g^2 + A^2} \]