The time period of oscillations of a simple pendulum is 1 minute. If its length is increased b 44% then its new time period of oscillation will be

Options

A. 96s
B. 58s
C. 82s
D. 72s

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the formula for the time period of a simple pendulum
The time period (T) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where:
- \( T \) is the time period,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

### Step 2: Identify the initial conditions
We are given that the initial time period \( T \) is 1 minute (or 60 seconds). Therefore:
\[ T = 60 \, \text{seconds} \]

### Step 3: Calculate the initial length of the pendulum
From the time period formula, we can express the length \( L \) in terms of \( T \):
\[ L = \frac{g T^2}{4\pi^2} \]

### Step 4: Increase the length by 44%
If the length is increased by 44%, the new length \( L' \) can be calculated as:
\[ L' = L + 0.44L = 1.44L \]

### Step 5: Calculate the new time period
Using the new length \( L' \), we can find the new time period \( T' \):
\[ T' = 2\pi \sqrt{\frac{L'}{g}} \]
Substituting \( L' \):
\[ T' = 2\pi \sqrt{\frac{1.44L}{g}} \]

### Step 6: Factor out the original time period
We can factor out the original time period from the equation:
\[ T' = 2\pi \sqrt{\frac{1.44L}{g}} = 2\pi \sqrt{1.44} \sqrt{\frac{L}{g}} \]
Since \( \sqrt{1.44} = 1.2 \), we have:
\[ T' = 1.2 \cdot 2\pi \sqrt{\frac{L}{g}} \]
Thus:
\[ T' = 1.2 T \]

### Step 7: Substitute the original time period
Now substituting \( T = 60 \) seconds:
\[ T' = 1.2 \cdot 60 = 72 \, \text{seconds} \]

### Conclusion
The new time period of oscillation after increasing the length by 44% is:
\[ T' = 72 \, \text{seconds} \]