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 analyze the situation when both masses are suspended and then when one mass is removed.

### Step 1: Understand the initial equilibrium condition
When both masses \( m_1 \) and \( m_2 \) are suspended, they reach an equilibrium position where the gravitational force acting downwards is balanced by the spring force acting upwards. The total weight of the system is given by:
\[ F_{\text{gravity}} = (m_1 + m_2)g \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 2: Write the equation for the spring force
The spring force can be expressed as:
\[ F_{\text{spring}} = kL \]
where \( k \) is the spring constant and \( L \) is the extension of the spring from its natural length.

### Step 3: Set up the equilibrium equation
At equilibrium, the forces balance:
\[ (m_1 + m_2)g = kL \]
Substituting the values:
- \( m_1 = 1 \, \text{kg} \)
- \( m_2 = 0.5 \, \text{kg} \)
- \( k = 12.5 \, \text{N/m} \)

The equation becomes:
\[ (1 + 0.5) \cdot 10 = 12.5L \]
\[ 15 = 12.5L \]
From this, we can find \( L \):
\[ L = \frac{15}{12.5} = 1.2 \, \text{m} \]

### Step 4: Analyze the situation after removing \( m_1 \)
Once \( m_1 \) is removed, only \( m_2 \) remains. The new equilibrium position will be determined by the weight of \( m_2 \):
\[ F_{\text{gravity}} = m_2 g = 0.5 \cdot 10 = 5 \, \text{N} \]

### Step 5: Write the new spring force equation
At the new equilibrium position, the spring force will now be:
\[ F_{\text{spring}} = kL' \]
where \( L' \) is the new extension of the spring.

### Step 6: Set up the new equilibrium equation
At the new equilibrium, we have:
\[ m_2 g = kL' \]
Substituting the values:
\[ 5 = 12.5L' \]
From this, we find \( L' \):
\[ L' = \frac{5}{12.5} = 0.4 \, \text{m} \]

### Step 7: Calculate the change in extension
The change in extension \( \Delta L \) (which will be the new amplitude of oscillation) is given by:
\[ \Delta L = L - L' = 1.2 - 0.4 = 0.8 \, \text{m} \]

### Step 8: Convert to centimeters
To convert the amplitude from meters to centimeters:
\[ \Delta L = 0.8 \, \text{m} \times 100 = 80 \, \text{cm} \]

### Final Answer
The new amplitude of oscillation after removing \( m_1 \) is:
\[ \Delta L = 80 \, \text{cm} \]