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 will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem:**
We are given that the tension in the string at the mean position of the pendulum is 1% more than the weight of the bob. We need to find the angular amplitude (θ) of the pendulum.

2. **Setting Up the Equation:**
Let the mass of the bob be \( m \) and the gravitational force (weight) acting on it be \( mg \). The tension \( T \) in the string can be expressed as:
\[ T = mg + 0.01 \cdot mg = mg(1 + 0.01) = 1.01mg \]

3. **Applying the Dynamics of the Pendulum:**
At the mean position, the tension can also be expressed in terms of the centripetal force:
\[ T = mg + \frac{mv^2}{L} \]
where \( L \) is the length of the string and \( v \) is the speed of the bob at the mean position.

4. **Equating the Two Expressions for Tension:**
Setting the two expressions for tension equal gives:
\[ 1.01mg = mg + \frac{mv^2}{L} \]
Simplifying this, we get:
\[ 0.01mg = \frac{mv^2}{L} \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ 0.01g = \frac{v^2}{L} \]
Rearranging gives:
\[ v^2 = 0.01gL \quad \text{(Equation 1)} \]

5. **Using Conservation of Energy:**
At the extreme position, the kinetic energy is zero and all energy is potential. The height \( h \) at the extreme position can be expressed as:
\[ h = L(1 - \cos \theta) \]
The potential energy at this height is:
\[ PE = mgh = mgL(1 - \cos \theta) \]
At the mean position, the kinetic energy is:
\[ KE = \frac{1}{2}mv^2 \]
By conservation of energy:
\[ mgL(1 - \cos \theta) = \frac{1}{2}mv^2 \]

6. **Substituting for \( v^2 \):**
From Equation 1, substitute \( v^2 \):
\[ mgL(1 - \cos \theta) = \frac{1}{2}m(0.01gL) \]
Simplifying gives:
\[ gL(1 - \cos \theta) = 0.005gL \]
Dividing both sides by \( gL \) (assuming \( gL \neq 0 \)):
\[ 1 - \cos \theta = 0.005 \]
Thus:
\[ \cos \theta = 1 - 0.005 = 0.995 \]

7. **Finding the Angular Amplitude:**
To find \( \theta \):
\[ \theta = \cos^{-1}(0.995) \]
Using a calculator, we find:
\[ \theta \approx 0.1 \text{ radians} \]

### Final Answer:
The angular amplitude of the pendulum is approximately \( 0.1 \) radians.

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