A ball of mass 2kg hanging from a spring oscillates with a time period `2pi` seconds. Ball is removed when it is in equilibrium position, then spring shortens by

Options

A. `10 m`
B. `5m`
C. `20m`
D. `2pim`

Answer

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

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

### Step 2: Substitute the given values
We know from the problem that the time period \( T \) is \( 2\pi \) seconds and the mass \( m \) is \( 2 \) kg. We can set up the equation:
\[ 2\pi = 2\pi \sqrt{\frac{2}{k}} \]

### Step 3: Simplify the equation
Dividing both sides by \( 2\pi \):
\[ 1 = \sqrt{\frac{2}{k}} \]
Now, squaring both sides gives:
\[ 1 = \frac{2}{k} \]

### Step 4: Solve for the spring constant \( k \)
Rearranging the equation to find \( k \):
\[ k = 2 \, \text{N/m} \]

### Step 5: Calculate the elongation of the spring at equilibrium
At equilibrium, the spring force equals the weight of the mass:
\[ kx_0 = mg \]
Substituting \( k = 2 \, \text{N/m} \), \( m = 2 \, \text{kg} \), and \( g = 10 \, \text{m/s}^2 \):
\[ 2x_0 = 2 \times 10 \]
This simplifies to:
\[ 2x_0 = 20 \]

### Step 6: Solve for \( x_0 \)
Dividing both sides by 2:
\[ x_0 = 10 \, \text{m} \]

### Step 7: Conclusion
When the ball is removed from the equilibrium position, the spring will shorten by \( 10 \, \text{m} \).

### Final Answer:
The spring shortens by \( 10 \, \text{m} \).

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