A body of mass `1kg` is suspended from a weightless spring having force constant `600N//m`. Another body of mass `0.5 kg` moving vertically upward hits the suspended body with a velocity of `3.0m//s` and get embedded in it. Find the frequency of oscillations and amplitude of motion.

Explanation

To solve the problem step by step, we will follow the principles of conservation of momentum and the formulas related to simple harmonic motion.

### Step 1: Understand the System
We have two masses:
- Mass \( m_1 = 1 \, \text{kg} \) (suspended from the spring)
- Mass \( m_2 = 0.5 \, \text{kg} \) (moving upward with a velocity of \( 3 \, \text{m/s} \))

### Step 2: Apply Conservation of Momentum
When mass \( m_2 \) collides with mass \( m_1 \), we can use the conservation of linear momentum to find the final velocity \( V' \) of the combined mass after the collision.

The initial momentum before the collision is:
\[ \text{Initial Momentum} = m_2 \cdot V + m_1 \cdot 0 = 0.5 \cdot 3 + 1 \cdot 0 = 1.5 \, \text{kg m/s} \]

The final momentum after the collision is:
\[ \text{Final Momentum} = (m_1 + m_2) \cdot V' = (1 + 0.5) \cdot V' = 1.5 \cdot V' \]

Setting the initial momentum equal to the final momentum:
\[ 1.5 = 1.5 \cdot V' \]
\[ V' = \frac{1.5}{1.5} = 1 \, \text{m/s} \]

### Step 3: Calculate the Frequency of Oscillation
The total mass \( M \) after the collision is:
\[ M = m_1 + m_2 = 1 + 0.5 = 1.5 \, \text{kg} \]

The spring constant \( k \) is given as \( 600 \, \text{N/m} \).

The angular frequency \( \omega \) is given by:
\[ \omega = \sqrt{\frac{k}{M}} = \sqrt{\frac{600}{1.5}} = \sqrt{400} = 20 \, \text{rad/s} \]

The frequency \( f \) is related to the angular frequency by:
\[ f = \frac{\omega}{2\pi} = \frac{20}{2\pi} = \frac{10}{\pi} \, \text{Hz} \]

### Step 4: Calculate the Amplitude of Motion
The amplitude \( A \) can be calculated using the formula:
\[ A = \frac{M \cdot V'}{k} \]
Substituting the values:
\[ A = \frac{1.5 \cdot 1}{600} = \frac{1.5}{600} = 0.0025 \, \text{m} = 2.5 \, \text{cm} \]

### Final Results
- Frequency of oscillation \( f = \frac{10}{\pi} \, \text{Hz} \)
- Amplitude of motion \( A = 2.5 \, \text{cm} \)