A ball of mass `m kg` hangs from a spring of spring constant `k`. The ball oscillates with a period of `T` seconds if the ball is removed, the spring is shortened(`w.r.t.` length in mean position) by

Options

A. `(gT^(2))/((2pi)^(2))` metre
B. `(3T^(2)g)/((2pi)^(2))` metre
C. `(Tm)/(k,)` metre
D. `(Tk)/(m)` metre

Answer

To solve the problem of how much the spring will be shortened when the ball is removed, we can follow these steps:

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

### Step 2: Rearrange the formula to find the spring constant \( k \)
We can rearrange the formula to express \( k \) in terms of \( T \) and \( m \):
\[ T^2 = 4\pi^2 \frac{m}{k} \]
This leads to:
\[ k = \frac{4\pi^2 m}{T^2} \]

### Step 3: Determine the forces acting on the spring when the ball is attached
When the ball of mass \( m \) is hanging from the spring, the force exerted by the ball due to gravity is:
\[ F_{\text{initial}} = mg \]
where \( g \) is the acceleration due to gravity.

### Step 4: Calculate the change in force when the ball is removed
When the ball is removed, the final force \( F_{\text{final}} \) acting on the spring becomes zero. The change in force \( \Delta F \) is given by:
\[ \Delta F = F_{\text{final}} - F_{\text{initial}} = 0 - mg = -mg \]

### Step 5: Relate the change in force to the change in displacement
Using Hooke's law, the change in displacement \( \Delta x \) can be expressed as:
\[ \Delta x = \frac{\Delta F}{-k} \]
Substituting for \( \Delta F \):
\[ \Delta x = \frac{-mg}{-k} = \frac{mg}{k} \]

### Step 6: Substitute the expression for \( k \) into the displacement equation
Now, we can substitute the expression for \( k \) that we derived earlier:
\[ \Delta x = \frac{mg}{\frac{4\pi^2 m}{T^2}} = \frac{mg T^2}{4\pi^2 m} \]
The mass \( m \) cancels out:
\[ \Delta x = \frac{g T^2}{4\pi^2} \]

### Final Answer
Thus, the spring will be shortened by:
\[ \Delta x = \frac{g T^2}{4\pi^2} \]

---