A second's pendulum is taken to a height where the value of g is `4.36ms^(-2)`. What will be its new time period?

Answer

To find the new time period of a second's pendulum when taken to a height where the value of g is 4.36 m/s², we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Time Period Formula**:
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.

2. **Identify the Time Period of a Second's Pendulum**:
A second's pendulum has a time period of 2 seconds at standard gravity (g = 9.8 m/s²). Thus, we can write:
\[ 2 = 2\pi \sqrt{\frac{l}{9.8}} \quad \text{(1)} \]

3. **Rearranging the Equation**:
From equation (1), we can isolate \( l \):
\[ 2 = 2\pi \sqrt{\frac{l}{9.8}} \implies 1 = \pi \sqrt{\frac{l}{9.8}} \implies \sqrt{\frac{l}{9.8}} = \frac{1}{\pi} \implies l = \frac{9.8}{\pi^2} \]

4. **Calculate the New Time Period**:
When the pendulum is taken to a height where \( g \) is 4.36 m/s², we can find the new time period \( T' \):
\[ T' = 2\pi \sqrt{\frac{l}{g'}} \quad \text{where } g' = 4.36 \text{ m/s}^2 \]
Substituting \( l \) from the previous step:
\[ T' = 2\pi \sqrt{\frac{\frac{9.8}{\pi^2}}{4.36}} \]

5. **Simplifying the Expression**:
\[ T' = 2\pi \sqrt{\frac{9.8}{4.36 \pi^2}} = 2\pi \cdot \frac{1}{\pi} \sqrt{\frac{9.8}{4.36}} = 2 \sqrt{\frac{9.8}{4.36}} \]

6. **Calculating the Numerical Value**:
Now we calculate \( \sqrt{\frac{9.8}{4.36}} \):
\[ \frac{9.8}{4.36} \approx 2.25 \implies \sqrt{2.25} = 1.5 \]
Therefore:
\[ T' = 2 \times 1.5 = 3 \text{ seconds} \]

### Final Answer:
The new time period of the pendulum at the height where \( g = 4.36 \, \text{m/s}^2 \) is approximately **3 seconds**.