A seconds pendulum is shifted from a place where `g = 9.8 m//s^(2)` to another place where `g = 9.78 m//s^(2)`. To keep period of oscillation constant its length should be

Options

A. decreased by `(2)/(pi^(2))cm`
B. increased by `(2)/(pi^(2))cm`
C. increased by `(2)/(pi)cm`
D. decreased by `(2)/(pi)cm`

Explanation

To solve the problem of determining how the length of a seconds pendulum should change when moving from a location with a gravitational acceleration of \( g_1 = 9.8 \, \text{m/s}^2 \) to another location with \( g_2 = 9.78 \, \text{m/s}^2 \) while keeping the period of oscillation constant, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for the Period of a 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. **Set Up the Equations for Two Locations:**
Since the period must remain constant when moving from one location to another, we can set up the equations for the two different gravitational accelerations:
\[ T_1 = 2\pi \sqrt{\frac{L_1}{g_1}} \quad \text{and} \quad T_2 = 2\pi \sqrt{\frac{L_2}{g_2}} \]
Since \( T_1 = T_2 \), we can equate the two:
\[ 2\pi \sqrt{\frac{L_1}{g_1}} = 2\pi \sqrt{\frac{L_2}{g_2}} \]

3. **Simplify the Equation:**
Dividing both sides by \( 2\pi \) gives:
\[ \sqrt{\frac{L_1}{g_1}} = \sqrt{\frac{L_2}{g_2}} \]
Squaring both sides results in:
\[ \frac{L_1}{g_1} = \frac{L_2}{g_2} \]

4. **Rearranging the Equation:**
Rearranging gives us:
\[ L_2 = L_1 \cdot \frac{g_2}{g_1} \]

5. **Substituting the Values of \( g_1 \) and \( g_2 \):**
Plugging in the values:
\[ g_1 = 9.8 \, \text{m/s}^2, \quad g_2 = 9.78 \, \text{m/s}^2 \]
We find:
\[ L_2 = L_1 \cdot \frac{9.78}{9.8} \]

6. **Calculating the Change in Length:**
The change in length \( \Delta L \) can be expressed as:
\[ \Delta L = L_2 - L_1 = L_1 \cdot \frac{9.78}{9.8} - L_1 = L_1 \left(\frac{9.78 - 9.8}{9.8}\right) \]
Simplifying gives:
\[ \Delta L = L_1 \left(-\frac{0.02}{9.8}\right) \]

7. **Interpreting the Result:**
Since \( \Delta L \) is negative, this indicates that the length \( L_2 \) is less than \( L_1 \), meaning the length must be decreased.

8. **Converting to Centimeters:**
To express the change in length in centimeters, we multiply by 100:
\[ \Delta L = -\frac{0.02}{9.8} \cdot L_1 \cdot 100 \]
This gives us the decrease in length in centimeters.

### Final Result:
The length of the pendulum should be decreased by approximately \( \frac{2}{\pi^2} \) centimeters to keep the period of oscillation constant.