A block of mass m is hung vertically from an elastic thread of force constant `mg//a`. Initially the thread was at its natural length and the block is allowed to fall freely. Kinetic energy of the block when it passes through the equilibrium position will be

Options

A. mga
B. `(mga)/(2)`
C. zero
D. 2mga

Answer

To solve the problem step by step, we will analyze the situation involving a block of mass \( m \) hung from an elastic thread with a force constant \( k = \frac{mg}{a} \). We need to find the kinetic energy of the block when it passes through the equilibrium position after being released from rest.

### Step-by-Step Solution:

1. **Identify the Forces at Equilibrium:**
At the equilibrium position, the forces acting on the block are:
- The weight of the block acting downwards: \( F_{\text{gravity}} = mg \)
- The restoring force of the elastic thread acting upwards: \( F_{\text{spring}} = kx \)

At equilibrium, these forces balance each other:
\[ mg = kx \]

2. **Substitute the Value of \( k \):**
Given that \( k = \frac{mg}{a} \), we can substitute this into the equilibrium equation:
\[ mg = \left(\frac{mg}{a}\right)x \]

3. **Solve for \( x \):**
Rearranging the equation to solve for \( x \):
\[ x = a \]
This means the block stretches the elastic thread by a distance \( a \) when it reaches the equilibrium position.

4. **Determine the Maximum Displacement (Amplitude):**
Since the block is released from the natural length of the thread, the maximum displacement (amplitude) \( A \) from the natural length to the equilibrium position is:
\[ A = a \]

5. **Calculate the Angular Frequency \( \omega \):**
The angular frequency \( \omega \) for a mass-spring system is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]
Substituting \( k = \frac{mg}{a} \):
\[ \omega = \sqrt{\frac{\frac{mg}{a}}{m}} = \sqrt{\frac{g}{a}} \]

6. **Find the Maximum Velocity \( V_{\text{max}} \):**
The maximum velocity \( V_{\text{max}} \) at the equilibrium position is given by:
\[ V_{\text{max}} = A \cdot \omega = a \cdot \sqrt{\frac{g}{a}} = \sqrt{ag} \]

7. **Calculate the Kinetic Energy at Equilibrium Position:**
The kinetic energy \( KE \) of the block when it passes through the equilibrium position is given by:
\[ KE = \frac{1}{2} m V_{\text{max}}^2 \]
Substituting \( V_{\text{max}} = \sqrt{ag} \):
\[ KE = \frac{1}{2} m (\sqrt{ag})^2 = \frac{1}{2} m (ag) = \frac{mga}{2} \]

### Final Answer:
The kinetic energy of the block when it passes through the equilibrium position is:
\[ KE = \frac{mga}{2} \]