The tension along a string at its mean position is 1% more than its weight. Find the angular amplitude of the pendulum (in radian)

Explanation

To solve the problem, we need to find the angular amplitude of a pendulum when the tension in the string at its mean position is 1% more than its weight. Let's break down the solution step by step.

### Step 1: Understand the Forces Acting on the Pendulum
At the mean position of the pendulum, the forces acting on the bob are:
- The weight of the bob (downward) = \( mg \)
- The tension in the string (upward) = \( T \)

According to the problem, the tension \( T \) is 1% more than the weight:
\[ T = mg + 0.01mg = 1.01mg \]

### Step 2: Apply Newton's Second Law
At the mean position, the net force acting on the bob is also equal to the centripetal force required to keep the bob moving in a circular path. Thus, we can write:
\[ T - mg = \frac{mv^2}{L} \]
Substituting the expression for tension:
\[ 1.01mg - mg = \frac{mv^2}{L} \]
This simplifies to:
\[ 0.01mg = \frac{mv^2}{L} \]

### Step 3: Cancel out the Mass
Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ 0.01g = \frac{v^2}{L} \]
Rearranging gives:
\[ v^2 = 0.01gL \]

### Step 4: Relate Velocity to Angular Amplitude
The maximum velocity \( v \) of the pendulum at the mean position can be expressed in terms of the angular amplitude \( \theta_{\text{max}} \) and the length of the pendulum \( L \):
\[ v = \omega A \]
Where \( A \) is the maximum displacement (amplitude) and \( \omega \) is the angular frequency. The amplitude can be expressed as:
\[ A = L \theta_{\text{max}} \]
Thus, we have:
\[ v = \omega L \theta_{\text{max}} \]

### Step 5: Substitute for Angular Frequency
The angular frequency \( \omega \) is related to the length \( L \) and acceleration due to gravity \( g \):
\[ \omega = \sqrt{\frac{g}{L}} \]
Substituting this into the expression for \( v \):
\[ v = \sqrt{\frac{g}{L}} L \theta_{\text{max}} = \sqrt{gL} \theta_{\text{max}} \]

### Step 6: Substitute Back into the Velocity Equation
Now we can substitute \( v^2 \) back into our earlier equation:
\[ 0.01gL = (\sqrt{gL} \theta_{\text{max}})^2 \]
This simplifies to:
\[ 0.01gL = gL \theta_{\text{max}}^2 \]

### Step 7: Solve for Angular Amplitude
Dividing both sides by \( gL \) (assuming \( gL \neq 0 \)):
\[ 0.01 = \theta_{\text{max}}^2 \]
Taking the square root gives:
\[ \theta_{\text{max}} = \sqrt{0.01} = 0.1 \text{ radians} \]

### Final Answer
The angular amplitude of the pendulum is:
\[ \theta_{\text{max}} = 0.1 \text{ radians} \]

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