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 follow the principles of equilibrium and simple harmonic motion (SHM).

### Step 1: Understand the Problem
We have a spring with a spring constant \( K = 2 \, \text{N/cm} \) and a mass \( m \) hanging from it. When an additional mass of \( 3 \, \text{kg} \) is placed on top of \( m \), we need to find the new amplitude of oscillation.

### Step 2: Convert Units
First, we need to convert the spring constant from N/cm to N/m:
\[ K = 2 \, \text{N/cm} = 2 \times 100 \, \text{N/m} = 200 \, \text{N/m} \]

### Step 3: Set Up the Equations
1. When the mass \( m \) is hanging in equilibrium, the force due to gravity is balanced by the spring force:
\[ mg = Kx_1 \]
where \( x_1 \) is the initial displacement of the spring.

2. When the additional mass \( 3 \, \text{kg} \) is added, the new equilibrium condition is:
\[ (m + 3)g = Kx_2 \]
where \( x_2 \) is the new displacement of the spring.

### Step 4: Subtract the Equations
To find the change in displacement due to the added mass, we subtract the first equation from the second:
\[ (m + 3)g - mg = Kx_2 - Kx_1 \]
This simplifies to:
\[ 3g = K(x_2 - x_1) \]

### Step 5: Solve for the Change in Displacement
Rearranging gives:
\[ x_2 - x_1 = \frac{3g}{K} \]

### Step 6: Substitute Values
Now, substitute \( g = 10 \, \text{m/s}^2 \) and \( K = 200 \, \text{N/m} \):
\[ x_2 - x_1 = \frac{3 \times 10}{200} \]
\[ x_2 - x_1 = \frac{30}{200} = 0.15 \, \text{m} \]

### Step 7: Conclusion
The new amplitude of oscillation after adding the mass is:
\[ \text{Amplitude} = 0.15 \, \text{m} \]

### Final Answer
The new amplitude of oscillation is \( 0.15 \, \text{m} \).

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