For a spring-mass system spring having spring constant 19.7 N/m is attached to it. What should be the value of mass m (approx.) so that it will give the same period as that of seconds pendulum?

Options

A. 1 kg
B. 2 kg
C. 3 kg
D. 4 kg

Answer

To solve the problem, we need to find the mass \( m \) that will give the same period as a seconds pendulum, which has a period of 2 seconds. We will use the formula for the time period of a spring-mass system and set it equal to the period of the seconds pendulum.

### Step-by-step Solution:

1. **Understand the time period of a spring-mass system**:
The formula for the time period \( T \) of a spring-mass system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass attached to the spring and \( k \) is the spring constant.

2. **Set the time period equal to that of a seconds pendulum**:
The time period of a seconds pendulum is 2 seconds. Therefore, we can set:
\[ 2 = 2\pi \sqrt{\frac{m}{k}} \]

3. **Simplify the equation**:
Divide both sides by 2:
\[ 1 = \pi \sqrt{\frac{m}{k}} \]

4. **Square both sides**:
Squaring both sides gives:
\[ 1 = \pi^2 \frac{m}{k} \]

5. **Rearrange to solve for \( m \)**:
Rearranging the equation gives:
\[ m = \frac{k}{\pi^2} \]

6. **Substitute the value of \( k \)**:
Given that the spring constant \( k = 19.7 \, \text{N/m} \):
\[ m = \frac{19.7}{\pi^2} \]

7. **Calculate the value of \( m \)**:
Using \( \pi \approx 3.14 \):
\[ \pi^2 \approx 9.86 \]
Therefore:
\[ m \approx \frac{19.7}{9.86} \approx 2.00 \, \text{kg} \]

Thus, the approximate value of mass \( m \) is **2 kg**.