The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude.

Explanation

To solve the problem, we need to find the angular amplitude of a pendulum given that the maximum tension in the string is double the minimum tension. Let's break down the solution step by step:

### Step 1: Understand the Forces Acting on the Pendulum
When the pendulum is at an angle θ, the forces acting on it are:
- The gravitational force (mg) acting downwards.
- The tension (T) in the string acting upwards along the string.

### Step 2: Write the Equation for Maximum Tension
At the lowest point (θ = 0°), the tension is maximum. The equation for maximum tension can be derived from the centripetal force requirement:
\[ T_{max} = mg + \frac{mv^2}{l} \]
where:
- \( T_{max} \) is the maximum tension,
- \( m \) is the mass of the pendulum,
- \( v \) is the velocity at the lowest point,
- \( l \) is the length of the pendulum.

### Step 3: Write the Equation for Minimum Tension
At the highest point (when the pendulum is at an angle θ), the tension is minimum. The equation for minimum tension is:
\[ T_{min} = mg \cos \theta \]
where:
- \( T_{min} \) is the minimum tension.

### Step 4: Relate Maximum and Minimum Tension
According to the problem, the maximum tension is double the minimum tension:
\[ T_{max} = 2T_{min} \]
Substituting the expressions for \( T_{max} \) and \( T_{min} \):
\[ mg + \frac{mv^2}{l} = 2(mg \cos \theta) \]

### Step 5: Simplify the Equation
We can simplify the equation:
\[ mg + \frac{mv^2}{l} = 2mg \cos \theta \]
Dividing through by \( m \):
\[ g + \frac{v^2}{l} = 2g \cos \theta \]

### Step 6: Use Conservation of Energy
At the lowest point, all energy is kinetic, and at the highest point, all energy is potential. Therefore:
\[ \frac{1}{2} mv^2 = mg h \]
where \( h \) is the height raised, given by:
\[ h = l(1 - \cos \theta) \]
Thus, we can write:
\[ \frac{1}{2} mv^2 = mg l(1 - \cos \theta) \]
Dividing by \( m \):
\[ \frac{1}{2} v^2 = g l(1 - \cos \theta) \]
So:
\[ v^2 = 2g l(1 - \cos \theta) \]

### Step 7: Substitute \( v^2 \) into the Tension Equation
Now substitute \( v^2 \) back into the tension equation:
\[ g + \frac{2g l(1 - \cos \theta)}{l} = 2g \cos \theta \]
This simplifies to:
\[ g + 2g(1 - \cos \theta) = 2g \cos \theta \]
\[ g + 2g - 2g \cos \theta = 0 \]
\[ 3g = 4g \cos \theta \]

### Step 8: Solve for \( \cos \theta \)
Rearranging gives:
\[ \cos \theta = \frac{3}{4} \]

### Step 9: Find the Angular Amplitude
Finally, we can find the angular amplitude:
\[ \theta = \cos^{-1}\left(\frac{3}{4}\right) \]

### Final Answer
The angular amplitude \( \theta \) is:
\[ \theta = \cos^{-1}\left(\frac{3}{4}\right) \]

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