A body of mass 1.0 kg is suspended from a weightless spring having force constant `600Nm^(-1)`. Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of `3.0ms^(-1)` and gets embedded in it. Find the frequency of oscillations and amplitude of motion.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the scenario
We have two bodies:
- Body 1 (mass \( m_1 = 1.0 \, \text{kg} \)) is suspended from a spring with a spring constant \( k = 600 \, \text{N/m} \).
- Body 2 (mass \( m_2 = 0.5 \, \text{kg} \)) is moving upwards with a velocity \( v = 3.0 \, \text{m/s} \) and collides with Body 1, becoming embedded in it.

### Step 2: Apply conservation of momentum
Before the collision, the momentum of the system is given by the momentum of Body 2 (since Body 1 is at rest):
\[ \text{Initial momentum} = m_2 \cdot v = 0.5 \, \text{kg} \cdot 3.0 \, \text{m/s} = 1.5 \, \text{kg m/s} \]

After the collision, both bodies move together with a combined mass \( m_1 + m_2 = 1.0 \, \text{kg} + 0.5 \, \text{kg} = 1.5 \, \text{kg} \) and a new velocity \( v' \):
\[ \text{Final momentum} = (m_1 + m_2) \cdot v' \]

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

Solving for \( v' \):
\[ v' = \frac{1.5 \, \text{kg m/s}}{1.5 \, \text{kg}} = 1.0 \, \text{m/s} \]

### Step 3: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) of the oscillation can be calculated using the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
where \( m \) is the total mass after the collision:
\[ \omega = \sqrt{\frac{600 \, \text{N/m}}{1.5 \, \text{kg}}} = \sqrt{400} = 20 \, \text{rad/s} \]

### Step 4: Calculate the frequency \( f \)
The frequency \( f \) is related to the angular frequency \( \omega \) by:
\[ f = \frac{\omega}{2\pi} \]
Substituting the value of \( \omega \):
\[ f = \frac{20 \, \text{rad/s}}{2\pi} = \frac{10}{\pi} \, \text{Hz} \]

### Step 5: Calculate the amplitude \( A \)
The amplitude \( A \) can be calculated using the formula:
\[ A = \frac{v'}{\omega} \]
Substituting the values of \( v' \) and \( \omega \):
\[ A = \frac{1.0 \, \text{m/s}}{20 \, \text{rad/s}} = 0.05 \, \text{m} = 5 \, \text{cm} \]

### Final Answers
- Frequency of oscillations: \( f = \frac{10}{\pi} \, \text{Hz} \)
- Amplitude of motion: \( A = 5 \, \text{cm} \)