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 osciallation is

Options

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

Answer

To solve the problem, we need to determine the amplitude of oscillation of the two blocks connected by a spring after one of the blocks is given an initial velocity. Here’s a step-by-step solution:

### Step 1: Understand the System
We have two blocks:
- Mass \( m_1 = 1 \, \text{kg} \)
- Mass \( m_2 = 2 \, \text{kg} \)
- Spring constant \( k = 24 \, \text{N/m} \)
- Initial velocity of \( m_1 \), \( v_0 = 12 \, \text{cm/s} = 0.12 \, \text{m} \)

### Step 2: Calculate the Reduced Mass
The reduced mass \( \mu \) of the system can be calculated using 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: Calculate the Initial Kinetic Energy
The initial kinetic energy (KE) of the system when \( m_1 \) is given an initial velocity \( v_0 \) is given by:
\[ KE = \frac{1}{2} \mu v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \cdot \frac{2}{3} \cdot (0.12)^2 \]
Calculating \( (0.12)^2 = 0.0144 \):
\[ KE = \frac{1}{2} \cdot \frac{2}{3} \cdot 0.0144 = \frac{1}{3} \cdot 0.0144 = 0.0048 \, \text{J} \]

### Step 4: Calculate the Maximum Potential Energy
At maximum compression of the spring, all the kinetic energy will convert into potential energy (PE) stored in the spring:
\[ PE = \frac{1}{2} k x^2 \]
Setting \( KE = PE \):
\[ 0.0048 = \frac{1}{2} \cdot 24 \cdot x^2 \]
This simplifies to:
\[ 0.0048 = 12 x^2 \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ x^2 = \frac{0.0048}{12} = 0.0004 \]
Taking the square root:
\[ x = \sqrt{0.0004} = 0.02 \, \text{m} = 2 \, \text{cm} \]

### Final Answer
The amplitude of oscillation is \( 2 \, \text{cm} \).

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