Two masses `m_(1) = 1kg and m_(2) = 0.5 kg` are suspended together by a massless spring of spring constant `12.5 Nm^(-1)`. When masses are in equilibrium `m_(1)` is removed without disturbing the system. New amplitude of oscillation will be

Options

A. 30cm
B. 50cm
C. 80cm
D. 60cm

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understand the system
We have two masses, \( m_1 = 1 \, \text{kg} \) and \( m_2 = 0.5 \, \text{kg} \), suspended by a spring with a spring constant \( k = 12.5 \, \text{N/m} \). Initially, both masses are in equilibrium.

### Step 2: Calculate the total weight acting on the spring
The total weight acting on the spring when both masses are present is given by:
\[ F_{\text{total}} = (m_1 + m_2) \cdot g = (1 \, \text{kg} + 0.5 \, \text{kg}) \cdot 10 \, \text{m/s}^2 = 15 \, \text{N} \]

### Step 3: Determine the extension of the spring at equilibrium
Using Hooke's Law, the extension \( x \) of the spring when both masses are attached can be calculated as:
\[ F_{\text{total}} = k \cdot x \implies x = \frac{F_{\text{total}}}{k} = \frac{15 \, \text{N}}{12.5 \, \text{N/m}} = 1.2 \, \text{m} \]

### Step 4: Remove mass \( m_1 \)
When mass \( m_1 \) is removed, the new weight acting on the spring is just \( m_2 \):
\[ F_{\text{new}} = m_2 \cdot g = 0.5 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 5 \, \text{N} \]

### Step 5: Calculate the new extension of the spring
Now, we calculate the new extension \( x' \) of the spring with only mass \( m_2 \):
\[ x' = \frac{F_{\text{new}}}{k} = \frac{5 \, \text{N}}{12.5 \, \text{N/m}} = 0.4 \, \text{m} \]

### Step 6: Determine the change in extension
The change in extension due to the removal of mass \( m_1 \) is:
\[ \Delta x = x - x' = 1.2 \, \text{m} - 0.4 \, \text{m} = 0.8 \, \text{m} \]

### Step 7: Find the new amplitude of oscillation
The new amplitude of oscillation for mass \( m_2 \) after removing \( m_1 \) is equal to the change in extension:
\[ \text{New Amplitude} = \Delta x = 0.8 \, \text{m} = 80 \, \text{cm} \]

### Conclusion
Thus, the new amplitude of oscillation when mass \( m_1 \) is removed is \( 80 \, \text{cm} \).

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