A spring of force constant k = 300 N/m connects two blocks having masses 2 kg and 3kg lying on a smooth horizontal plane. If the spring block system is released from a stretched position find the number of complete oscillations in 1 minute. Take `pi = sqrt10`

Answer

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

### Step 1: Identify the effective mass of the system
The effective mass \( \mu \) of two blocks connected by a spring can be calculated using the formula:
\[ \mu = \frac{m_1 \cdot m_2}{m_1 + m_2} \]
where \( m_1 = 2 \, \text{kg} \) and \( m_2 = 3 \, \text{kg} \).

### Step 2: Calculate the effective mass
Substituting the values into the formula:
\[ \mu = \frac{2 \cdot 3}{2 + 3} = \frac{6}{5} \, \text{kg} \]

### Step 3: Write the formula for the time period of oscillation
The time period \( T \) of the spring-block system is given by:
\[ T = 2\pi \sqrt{\frac{\mu}{k}} \]
where \( k = 300 \, \text{N/m} \).

### Step 4: Substitute the effective mass and spring constant into the time period formula
Substituting \( \mu \) and \( k \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{6}{5}}{300}} = 2\pi \sqrt{\frac{6}{1500}} = 2\pi \sqrt{\frac{1}{250}} = 2\pi \cdot \frac{1}{\sqrt{250}} \]

### Step 5: Simplify the square root
We know that \( \sqrt{250} = \sqrt{25 \cdot 10} = 5\sqrt{10} \). Thus:
\[ T = 2\pi \cdot \frac{1}{5\sqrt{10}} = \frac{2\pi}{5\sqrt{10}} \]

### Step 6: Substitute \( \pi = \sqrt{10} \)
Now substituting \( \pi = \sqrt{10} \):
\[ T = \frac{2\sqrt{10}}{5\sqrt{10}} = \frac{2}{5} \, \text{seconds} \]

### Step 7: Calculate the frequency
The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{1}{\frac{2}{5}} = \frac{5}{2} \, \text{Hz} \]

### Step 8: Find the number of complete oscillations in one minute
To find the number of complete oscillations in one minute, we multiply the frequency by 60 seconds:
\[ \text{Number of oscillations in 1 minute} = f \times 60 = \frac{5}{2} \times 60 = 150 \]

### Final Answer
The number of complete oscillations in one minute is **150**.

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