A block of mass 1 kg is connected with a smooth plank of the same mass is performing oscillations. The value of the spring constant is `200 N m^(-1)` The block and the plank are free to move and there is no friction anywhere. The angular frequency of the oscillation is `omega` rad `s^(-1)` Find the value of `omega`

Answer

To find the angular frequency \( \omega \) of the oscillating system consisting of a block and a plank, we can follow these steps:

### Step 1: Identify the parameters
We have:
- Mass of the block, \( m = 1 \, \text{kg} \)
- Mass of the plank, \( m = 1 \, \text{kg} \)
- Spring constant, \( k = 200 \, \text{N/m} \)

### Step 2: Understand the system
The system consists of two masses (the block and the plank) connected by a spring. Since there is no friction, both masses can move freely.

### Step 3: Use the concept of reduced mass
For a system with two masses \( m_1 \) and \( m_2 \), the reduced mass \( \mu \) is given by:
\[ \mu = \frac{m_1 \cdot m_2}{m_1 + m_2} \]
In our case, both masses are equal:
\[ m_1 = m_2 = 1 \, \text{kg} \]
Thus, the reduced mass \( \mu \) becomes:
\[ \mu = \frac{1 \cdot 1}{1 + 1} = \frac{1}{2} \, \text{kg} = 0.5 \, \text{kg} \]

### Step 4: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) for a mass-spring system is given by:
\[ \omega = \sqrt{\frac{k}{\mu}} \]
Substituting the values of \( k \) and \( \mu \):
\[ \omega = \sqrt{\frac{200}{0.5}} = \sqrt{400} \]

### Step 5: Find the final value of \( \omega \)
Calculating the square root:
\[ \omega = 20 \, \text{rad/s} \]

### Conclusion
Thus, the angular frequency of the oscillation is:
\[ \omega = 20 \, \text{rad/s} \]

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