Two blocks of masses `m_1`=1kg and `m_2` = 2kg are connected by a spring of spring constant k = 24 N/m and placed on a frictionless horizontal surface. The block `m_(1)` is imparted an initial velocity `v_(0)` = 12cm/s to the right. The amplitude of oscillation is

Options

A. 2 cm
B. 1 cm
C. 3 cm
D. 4 cm

Answer

To find the amplitude of oscillation for the two blocks connected by a spring, we can follow these steps:

### Step 1: Understand the System
We have two blocks with masses:
- \( m_1 = 1 \, \text{kg} \)
- \( m_2 = 2 \, \text{kg} \)

These blocks are connected by a spring with a spring constant \( k = 24 \, \text{N/m} \). The block \( m_1 \) is given an initial velocity \( v_0 = 12 \, \text{cm/s} = 0.12 \, \text{m/s} \) to the right.

### Step 2: Calculate the Reduced Mass
The reduced mass \( \mu \) for two masses \( m_1 \) and \( m_2 \) is given by the formula:
\[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]
Substituting the values:
\[ \mu = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \, \text{kg} \]

### Step 3: Use Conservation of Mechanical Energy
The total mechanical energy in the system is conserved. At the mean position (where the spring is neither compressed nor extended), all energy is kinetic:
\[ E = \frac{1}{2} \mu v_0^2 \]
At maximum compression (where the spring is fully compressed), all energy is potential:
\[ E = \frac{1}{2} k a^2 \]
Setting these equal gives:
\[ \frac{1}{2} \mu v_0^2 = \frac{1}{2} k a^2 \]

### Step 4: Substitute Known Values
Substituting the known values into the energy conservation equation:
\[ \frac{1}{2} \left(\frac{2}{3}\right) (0.12)^2 = \frac{1}{2} (24) a^2 \]
This simplifies to:
\[ \frac{1}{3} (0.12)^2 = 24 a^2 \]

### Step 5: Calculate \( v_0^2 \)
Calculating \( (0.12)^2 \):
\[ (0.12)^2 = 0.0144 \]
Thus,
\[ \frac{1}{3} \times 0.0144 = 24 a^2 \]
\[ 0.0048 = 24 a^2 \]

### Step 6: Solve for \( a^2 \)
To find \( a^2 \):
\[ a^2 = \frac{0.0048}{24} = 0.0002 \]

### Step 7: Find Amplitude \( a \)
Taking the square root to find \( a \):
\[ a = \sqrt{0.0002} = 0.01414 \, \text{m} = 1.414 \, \text{cm} \]

### Step 8: Convert to Centimeters
Since the question asks for the amplitude in centimeters:
\[ a \approx 1.41 \, \text{cm} \]

### Final Answer
The amplitude of oscillation is approximately \( 1.41 \, \text{cm} \).

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