The maximum tension in the string of a pendulum is three times the minimum tension:

Options

A. Maximum tension in string is `(4mg)/(5)`
B. Maximum tension in string is `(6mg)/(5)`
C. Maximum speed of bob is `3sqrt((gl)/(5))`
D. Maximum speed of bob is `2sqrt((gl)/(5))`

Answer

To solve the problem, we need to analyze the forces acting on a pendulum at different positions and derive the relationship between maximum and minimum tension in the string.

### Step-by-Step Solution:

1. **Understanding the Pendulum Position**:
- The maximum tension in the string occurs when the pendulum is at its lowest point (vertical position).
- The minimum tension occurs when the pendulum is at its extreme position (maximum angle from the vertical).

2. **Forces Acting on the Pendulum**:
- At the lowest point (maximum tension), the forces acting on the pendulum bob are:
- Weight (mg) acting downwards.
- Tension (T_max) acting upwards.
- At the extreme position (minimum tension), the forces are:
- Weight (mg) acting downwards.
- Tension (T_min) acting along the string.

3. **Expression for Tension**:
- At the lowest point:
\[ T_{max} = mg + \frac{mv^2}{L} \]
- At the extreme position:
\[ T_{min} = mg \cos \theta \]
- Here, \(v\) is the velocity at the lowest point, and \(L\) is the length of the string.

4. **Given Condition**:
- We are given that the maximum tension is three times the minimum tension:
\[ T_{max} = 3 \cdot T_{min} \]

5. **Substituting the Expressions**:
- Substitute the expressions for \(T_{max}\) and \(T_{min}\):
\[ mg + \frac{mv^2}{L} = 3(mg \cos \theta) \]

6. **Using Work-Energy Principle**:
- The work done by gravity when the pendulum swings from the extreme position to the lowest point is equal to the change in kinetic energy.
- The height change is \(L - L \cos \theta = L(1 - \cos \theta)\).
- Work done by gravity:
\[ W = mgL(1 - \cos \theta) \]
- Change in kinetic energy:
\[ \Delta KE = \frac{1}{2} mv^2 \]
- Setting the work done equal to the change in kinetic energy:
\[ mgL(1 - \cos \theta) = \frac{1}{2} mv^2 \]

7. **Solving for v²**:
- Cancel \(m\) from both sides:
\[ gL(1 - \cos \theta) = \frac{1}{2} v^2 \]
- Rearranging gives:
\[ v^2 = 2gL(1 - \cos \theta) \]

8. **Substituting v² back into T_max equation**:
- Substitute \(v^2\) into the \(T_{max}\) equation:
\[ T_{max} = mg + \frac{m(2gL(1 - \cos \theta))}{L} \]
- Simplifying gives:
\[ T_{max} = mg + 2mg(1 - \cos \theta) = mg(3 - 2\cos \theta) \]

9. **Equating T_max and T_min**:
- From the earlier equation:
\[ mg(3 - 2\cos \theta) = 3(mg \cos \theta) \]
- Cancel \(mg\):
\[ 3 - 2\cos \theta = 3\cos \theta \]
- Rearranging gives:
\[ 3 = 5\cos \theta \implies \cos \theta = \frac{3}{5} \]

10. **Conclusion**:
- The relationship between maximum and minimum tension in the pendulum string is established, and we find that the angle \(\theta\) satisfies \(\cos \theta = \frac{3}{5}\).