A 100 g mass stretches a particular spring by 9.8 cm, when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be 6.28 s?

Options

A. 1000g
B. `10^(5)g`
C. `10^(7)g`
D. `10^(4)g`

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Understand the given data
We are given:
- A mass \( m_1 = 100 \, \text{g} = 0.1 \, \text{kg} \) stretches the spring by \( x_0 = 9.8 \, \text{cm} = 0.098 \, \text{m} \).
- We need to find the mass \( m_2 \) that will give a period of vibration \( T = 6.28 \, \text{s} \).

### Step 2: Calculate the spring constant \( k \)
Using the balance of forces when the mass is at equilibrium:
\[ k x_0 = m_1 g \]
Where \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).

Rearranging the equation to find \( k \):
\[ k = \frac{m_1 g}{x_0} \]
Substituting the values:
\[ k = \frac{0.1 \times 9.8}{0.098} = \frac{0.98}{0.098} = 10 \, \text{N/m} \]

### Step 3: Use the formula for the period of oscillation
The period of a spring-mass system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
We need to rearrange this to find the mass \( m \):
\[ \frac{T}{2\pi} = \sqrt{\frac{m}{k}} \]
Squaring both sides:
\[ \left(\frac{T}{2\pi}\right)^2 = \frac{m}{k} \]
Thus,
\[ m = k \left(\frac{T}{2\pi}\right)^2 \]

### Step 4: Substitute the values to find \( m_2 \)
Now substituting \( k = 10 \, \text{N/m} \) and \( T = 6.28 \, \text{s} \):
\[ m_2 = 10 \left(\frac{6.28}{2\pi}\right)^2 \]
Calculating \( \frac{6.28}{2\pi} \):
\[ \frac{6.28}{2\pi} \approx 1 \]
Thus,
\[ m_2 = 10 \times 1^2 = 10 \, \text{kg} \]

### Step 5: Convert to grams
Since the question asks for mass in grams:
\[ m_2 = 10 \, \text{kg} = 10000 \, \text{g} \]

### Conclusion
The mass that must be attached to the spring for the period of vibration to be 6.28 seconds is \( 10000 \, \text{g} \).

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