The period of oscillation of simple pendulum increases by 20 % , when its length is increased by 44 cm. find its initial length.

Answer

To solve the problem, we need to find the initial length \( L_1 \) of the pendulum given that the period of oscillation increases by 20% when the length is increased by 44 cm.

### Step-by-Step Solution:

1. **Understanding the Period of a Simple Pendulum**:
The period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Setting Up the Initial and Final Periods**:
Let the initial length of the pendulum be \( L_1 \) and the initial period be \( T_1 \):
\[ T_1 = 2\pi \sqrt{\frac{L_1}{g}} \]
When the length is increased by 44 cm (0.44 m), the new length \( L_2 \) becomes:
\[ L_2 = L_1 + 0.44 \]
The new period \( T_2 \) becomes:
\[ T_2 = 2\pi \sqrt{\frac{L_2}{g}} = 2\pi \sqrt{\frac{L_1 + 0.44}{g}} \]

3. **Relating the Periods**:
According to the problem, the period increases by 20%, so:
\[ T_2 = T_1 + 0.2 T_1 = 1.2 T_1 \]
Therefore, we can write:
\[ 2\pi \sqrt{\frac{L_1 + 0.44}{g}} = 1.2 \cdot 2\pi \sqrt{\frac{L_1}{g}} \]
Dividing both sides by \( 2\pi \):
\[ \sqrt{\frac{L_1 + 0.44}{g}} = 1.2 \sqrt{\frac{L_1}{g}} \]

4. **Squaring Both Sides**:
Squaring both sides to eliminate the square root gives:
\[ \frac{L_1 + 0.44}{g} = 1.44 \cdot \frac{L_1}{g} \]
Since \( g \) is common on both sides, we can cancel it out:
\[ L_1 + 0.44 = 1.44 L_1 \]

5. **Rearranging the Equation**:
Rearranging the equation gives:
\[ 0.44 = 1.44 L_1 - L_1 \]
\[ 0.44 = 0.44 L_1 \]

6. **Solving for \( L_1 \)**:
Dividing both sides by 0.44:
\[ L_1 = 1 \text{ m} \]

### Final Answer:
The initial length of the pendulum \( L_1 \) is **1 meter**.