A disc of mass 3 m and a disc of mass m are connected by a massless spring of stiffness k. The heavier disc is placed on the ground with the spring vertical and lighter disc on top from its equilibrium position the upper disc is pushed down by a distance `delta` and released. Then.

Options

A. if `deltagt(3mg)/k` the lower disc will bounce up.
B. if `delta=(2 mg)/k` maximum normal reaction on lower disc = 6mg .
C. if `delta=(2 mg)/k` maximum normal reaction from ground on lower disc = 4 mg.
D. if `delta gt (4 mg)/k`, the lower disc will bounce up

Explanation

To solve the problem step by step, we will analyze the system of two discs connected by a spring and determine the conditions under which the lower disc will bounce up, as well as the maximum and minimum normal reactions acting on the lower disc.

### Step 1: Understand the System
We have two discs:
- A heavier disc of mass \(3m\) placed on the ground.
- A lighter disc of mass \(m\) placed on top of the heavier disc, connected by a spring of stiffness \(k\).

### Step 2: Determine the Equilibrium Position
In the equilibrium position, the weight of the upper disc (\(mg\)) is balanced by the spring force. Let the compression of the spring at equilibrium be \(\delta_0\).

Using Hooke's Law:
\[ F_{\text{spring}} = k \delta_0 = mg \]
From this, we can find the equilibrium compression:
\[ \delta_0 = \frac{mg}{k} \]

### Step 3: Analyze the Initial Displacement
The upper disc is pushed down by a distance \(\delta\) from its equilibrium position and then released. The total compression of the spring when the upper disc is pushed down is:
\[ x = \delta_0 + \delta = \frac{mg}{k} + \delta \]

### Step 4: Condition for Lower Disc to Bounce Up
The lower disc will bounce up if the force exerted by the spring exceeds the weight of the lower disc. The force exerted by the spring when compressed by \(x\) is:
\[ F_{\text{spring}} = kx = k\left(\frac{mg}{k} + \delta\right) = mg + k\delta \]
For the lower disc to bounce up:
\[ F_{\text{spring}} > 3mg \]
This leads to the condition:
\[ mg + k\delta > 3mg \implies k\delta > 2mg \implies \delta > \frac{2mg}{k} \]

### Step 5: Maximum Normal Reaction on the Lower Disc
If \(\delta = \frac{2mg}{k}\), we can calculate the maximum normal reaction (\(N_{\text{max}}\)) on the lower disc. The total downward force on the lower disc is:
\[ F_{\text{down}} = 3mg + F_{\text{spring}} = 3mg + 3mg = 6mg \]
Thus, the maximum normal reaction is:
\[ N_{\text{max}} = 6mg \]

### Step 6: Minimum Normal Reaction on the Lower Disc
If \(\delta = \frac{2mg}{k}\), we can also find the minimum normal reaction. The compression of the spring at this point is:
\[ x = \frac{mg}{k} + \frac{2mg}{k} = \frac{3mg}{k} \]
The force exerted by the spring is:
\[ F_{\text{spring}} = kx = 3mg \]
The minimum normal reaction occurs when the spring force just balances the weight of the upper disc:
\[ N_{\text{min}} = 3mg - mg = 2mg \]

### Step 7: Condition for Lower Disc to Bounce Up Again
If \(\delta > \frac{4mg}{k}\), we can analyze the forces again. The spring force must be equal to or greater than the weight of the lower disc for it to bounce up:
\[ F_{\text{spring}} = 3mg \]
Thus, the condition is:
\[ k\delta > 3mg \implies \delta > \frac{3mg}{k} \]

### Conclusion
- The lower disc will bounce up if \(\delta > \frac{3mg}{k}\).
- The maximum normal reaction on the lower disc when \(\delta = \frac{2mg}{k}\) is \(6mg\).
- The minimum normal reaction on the lower disc when \(\delta = \frac{2mg}{k}\) is \(2mg\).
- The lower disc will bounce up again if \(\delta > \frac{4mg}{k}\).