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 of finding the mass \( m \) for a spring-mass system to have the same period as a seconds pendulum, we can follow these steps:

### Step 1: Understand the Time Period of the Seconds Pendulum
The time period \( T \) of a seconds pendulum is 2 seconds.

### Step 2: Write the Formula for the Time Period of a Spring-Mass System
The time period \( T \) of a spring-mass system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where:
- \( m \) is the mass attached to the spring,
- \( k \) is the spring constant.

### Step 3: Set the Time Period of the Spring-Mass System Equal to that of the Seconds Pendulum
Since we want the time period of the spring-mass system to be equal to that of the seconds pendulum, we can set:
\[ 2\pi \sqrt{\frac{m}{k}} = 2 \]

### Step 4: Simplify the Equation
Dividing both sides by 2:
\[ \pi \sqrt{\frac{m}{k}} = 1 \]

### Step 5: Isolate the Square Root
Now, we can isolate the square root:
\[ \sqrt{\frac{m}{k}} = \frac{1}{\pi} \]

### Step 6: Square Both Sides
Next, we square both sides to eliminate the square root:
\[ \frac{m}{k} = \frac{1}{\pi^2} \]

### Step 7: Solve for Mass \( m \)
Now, we can solve for mass \( m \):
\[ m = k \cdot \frac{1}{\pi^2} \]

### Step 8: Substitute the Given Spring Constant
Given that the spring constant \( k = 19.7 \, \text{N/m} \):
\[ m = 19.7 \cdot \frac{1}{\pi^2} \]

### Step 9: Calculate \( \pi^2 \)
Using the approximation \( \pi \approx 3.14 \):
\[ \pi^2 \approx 9.86 \]

### Step 10: Final Calculation for Mass \( m \)
Now substituting \( \pi^2 \) into the equation:
\[ m \approx \frac{19.7}{9.86} \approx 1.997 \, \text{kg} \]

### Step 11: Round to the Nearest Whole Number
Rounding \( 1.997 \) gives approximately \( 2 \, \text{kg} \).

### Conclusion
Thus, the mass \( m \) that will give the same period as that of a seconds pendulum is approximately \( 2 \, \text{kg} \).