When a mass of 1 kg is suspended from a spring, it is stretched by 0.4m. A mass of 0.25 kg is suspended from the spring and the spring is allowed to oscillate. If `g=10 m//s^(2)`, then its period of oscillation will be

Options

A. 0.5 sec
B. 0.4 sec
C. 0.628 sec
D. 1.5 sec

Explanation

To solve the problem, we need to find the period of oscillation of a mass-spring system. Here are the steps to arrive at the solution:

### Step 1: Understand the problem and gather the given data
- We have a mass \( M = 1 \, \text{kg} \) that stretches the spring by \( x = 0.4 \, \text{m} \).
- We also have a smaller mass \( m = 0.25 \, \text{kg} \) that will oscillate on the same spring.
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).

### Step 2: Calculate the spring constant \( k \)
The weight of the mass \( M \) causes the spring to stretch. The force exerted by the mass is given by:
\[ F = M \cdot g = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10 \, \text{N} \]
According to Hooke's Law, the force exerted by the spring is:
\[ F = k \cdot x \]
Setting the two expressions for force equal gives:
\[ k \cdot x = M \cdot g \]
Substituting the known values:
\[ k \cdot 0.4 = 10 \]
Solving for \( k \):
\[ k = \frac{10}{0.4} = 25 \, \text{N/m} \]

### Step 3: Use the formula for the period of oscillation
The formula for the period \( T \) of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Where \( m \) is the mass that is oscillating (in this case, \( m = 0.25 \, \text{kg} \)) and \( k \) is the spring constant we just calculated.

### Step 4: Substitute the values into the formula
Substituting \( m = 0.25 \, \text{kg} \) and \( k = 25 \, \text{N/m} \):
\[ T = 2\pi \sqrt{\frac{0.25}{25}} \]
Calculating the fraction:
\[ \frac{0.25}{25} = 0.01 \]
Now, taking the square root:
\[ \sqrt{0.01} = 0.1 \]
Thus, substituting back into the formula:
\[ T = 2\pi \cdot 0.1 = 0.2\pi \]

### Step 5: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ T \approx 0.2 \cdot 3.14 = 0.628 \, \text{s} \]

### Final Answer
The period of oscillation is approximately \( 0.628 \, \text{s} \).