A simple pendulum swings with angular amplitude `theta`. The tension in the string when it is vertical is twice the tension in its extreme position. Then, ` cos theta` is equal to

Answer

To solve the problem, we need to analyze the forces acting on the pendulum at two positions: the extreme position (where the pendulum is at its maximum angular displacement, θ) and the vertical position (where the pendulum is at its lowest point).

### Step-by-Step Solution:

1. **Identify the Forces at the Vertical Position:**
At the vertical position, the tension in the string (T1) must counteract the weight of the pendulum bob (mg) and provide the necessary centripetal force for the circular motion. The equation for the forces at this position is:
\[ T_1 - mg = \frac{mu^2}{L} \]
where \( u \) is the speed of the bob at the vertical position and \( L \) is the length of the pendulum.

2. **Identify the Forces at the Extreme Position:**
At the extreme position, the pendulum is momentarily at rest, so the tension (T2) only needs to balance the weight of the bob:
\[ T_2 = mg \cos \theta \]
Here, the angle θ is the maximum angular displacement from the vertical.

3. **Relate T1 and T2:**
According to the problem, the tension at the vertical position is twice that at the extreme position:
\[ T_1 = 2T_2 \]

4. **Substitute T2 into the Equation:**
From the equation for T2, we can substitute:
\[ T_1 = 2(mg \cos \theta) \]

5. **Substituting T1 in the Force Equation:**
Now we substitute T1 in the first equation:
\[ 2(mg \cos \theta) - mg = \frac{mu^2}{L} \]
Simplifying this gives:
\[ mg(2 \cos \theta - 1) = \frac{mu^2}{L} \]

6. **Using Conservation of Energy:**
The potential energy lost when the pendulum moves from the extreme position to the vertical position is equal to the kinetic energy at the vertical position. The height (h) the bob descends is:
\[ h = L - L \cos \theta = L(1 - \cos \theta) \]
Therefore, the work done by gravity is:
\[ mg h = mg L(1 - \cos \theta) \]
This work done is equal to the change in kinetic energy:
\[ mg L(1 - \cos \theta) = \frac{1}{2} mu^2 \]

7. **Equating the Two Expressions:**
From the previous steps, we have:
\[ mg(2 \cos \theta - 1) = \frac{mu^2}{L} \]
and
\[ mg L(1 - \cos \theta) = \frac{1}{2} mu^2 \]
Now, we can substitute \( \frac{u^2}{L} \) from the second equation into the first equation.

8. **Solve for cos θ:**
By equating and simplifying, we can find:
\[ 2(2 \cos \theta - 1) = 1 - \cos \theta \]
Rearranging gives:
\[ 4 \cos \theta - 2 = 1 - \cos \theta \]
\[ 5 \cos \theta = 3 \]
\[ \cos \theta = \frac{3}{4} \]

### Final Answer:
\[ \cos \theta = \frac{3}{4} \]