When a mass of 5 kg is suspended from a spring of negligible mass and spring constant K, it oscillates with a periodic time `2pi`. If the mass is removed, the length of the spring will decrease by

Options

A. glk metre
B. klg metre
C. `2pi` metre
D. g metre

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the problem
We have a mass of 5 kg suspended from a spring with a spring constant \( K \). The mass oscillates with a periodic time of \( 2\pi \). We need to find out how much the length of the spring decreases when the mass is removed.

### Step 2: Use the formula for the time period of oscillation
The time period \( T \) of a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{K}} \]
Given that \( T = 2\pi \), we can set up the equation:
\[ 2\pi = 2\pi \sqrt{\frac{m}{K}} \]

### Step 3: Simplify the equation
Dividing both sides by \( 2\pi \):
\[ 1 = \sqrt{\frac{m}{K}} \]
Squaring both sides gives:
\[ 1 = \frac{m}{K} \]
Rearranging this, we find:
\[ K = m \]
Substituting the mass \( m = 5 \) kg:
\[ K = 5 \text{ N/m} \]

### Step 4: Apply Hooke's Law at equilibrium
At equilibrium, the weight of the mass is balanced by the spring force:
\[ mg = Kx_0 \]
Where \( x_0 \) is the elongation of the spring. Substituting the values:
\[ 5g = 5x_0 \]
Dividing both sides by 5:
\[ g = x_0 \]

### Step 5: Solve for \( x_0 \)
The acceleration due to gravity \( g \) is approximately \( 9.8 \, \text{m/s}^2 \). Therefore:
\[ x_0 = g \approx 9.8 \, \text{m} \]

### Step 6: Conclusion
When the mass is removed, the spring will return to its natural length, and the decrease in length will be equal to the elongation \( x_0 \). Thus, the length of the spring will decrease by approximately \( 9.8 \, \text{m} \).

### Final Answer
The length of the spring will decrease by \( g \) meters, which is approximately \( 9.8 \, \text{m} \).

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