A mass hangs in equilibrium from a spring of constant `K=2N//cm`Another mass of 3 kg is placed over M. Find the new amplitude of oscillation after wards (in m) `(Take g=10 m// s^(-2))`

Explanation

To solve the problem step by step, we will analyze the forces acting on the mass hanging from the spring and determine the new amplitude of oscillation after adding an additional mass.

### Step 1: Understand the initial equilibrium condition
When a mass \( M \) hangs from a spring with spring constant \( K \), it stretches the spring by a distance \( x_1 \) such that the force due to gravity on the mass equals the spring force.

The force due to gravity is given by:
\[ F_g = Mg \]
where \( g = 10 \, \text{m/s}^2 \).

The spring force is given by:
\[ F_s = Kx_1 \]

At equilibrium:
\[ Mg = Kx_1 \quad \text{(Equation 1)} \]

### Step 2: Introduce the additional mass
Now, when an additional mass of \( 3 \, \text{kg} \) is placed on the original mass \( M \), the new total mass becomes \( M + 3 \, \text{kg} \). The new displacement from the equilibrium position becomes \( x_2 \).

The new force due to gravity is:
\[ F_g' = (M + 3)g \]

The spring force at the new displacement is:
\[ F_s' = Kx_2 \]

At the new equilibrium:
\[ (M + 3)g = Kx_2 \quad \text{(Equation 2)} \]

### Step 3: Set up the equations
From Equations 1 and 2, we have:
1. \( Mg = Kx_1 \)
2. \( (M + 3)g = Kx_2 \)

### Step 4: Subtract the first equation from the second
Subtract Equation 1 from Equation 2 to find the difference in displacement:
\[ (M + 3)g - Mg = Kx_2 - Kx_1 \]
This simplifies to:
\[ 3g = K(x_2 - x_1) \]

### Step 5: Solve for the amplitude
The difference \( x_2 - x_1 \) represents the change in displacement, which is the amplitude of oscillation \( A \):
\[ A = x_2 - x_1 = \frac{3g}{K} \]

### Step 6: Substitute the values
Given:
- \( g = 10 \, \text{m/s}^2 \)
- \( K = 2 \, \text{N/cm} = 200 \, \text{N/m} \) (since \( 1 \, \text{cm} = 0.01 \, \text{m} \))

Substituting these values into the amplitude equation:
\[ A = \frac{3 \times 10}{200} = \frac{30}{200} = 0.15 \, \text{m} \]

### Final Answer:
The new amplitude of oscillation after adding the mass is:
\[ \boxed{0.15 \, \text{m}} \]