Two masses M and m are suspended together by massless spring of force constant -k. When the masses are in equilibrium, M is removed without disturbing the system. The amplitude of oscillations.

Options

A. `(Mg)/k`
B. `(mg)/k`
C. `((M+m)g)/k`
D. `((M-m)g)/k`

Answer

To solve the problem step by step, we will analyze the situation with the two masses suspended by a spring and how the removal of one mass affects the oscillations of the system.

### Step 1: Understand the Initial Setup
We have two masses, \( M \) and \( m \), suspended from a spring with a spring constant \( k \). When both masses are in equilibrium, the spring stretches due to the combined weight of both masses.

**Hint:** Identify the forces acting on the system when both masses are attached.

### Step 2: Determine the Initial Equilibrium Position
The force due to gravity acting on the system is \( (M + m)g \). At equilibrium, the spring force equals the gravitational force:
\[ k x_0 = (M + m)g \]
where \( x_0 \) is the initial elongation of the spring.

**Hint:** Use the equilibrium condition to relate the spring force and the gravitational force.

### Step 3: Solve for Initial Elongation \( x_0 \)
From the equilibrium condition, we can solve for \( x_0 \):
\[ x_0 = \frac{(M + m)g}{k} \]

**Hint:** Rearrange the equation to isolate \( x_0 \).

### Step 4: Remove Mass \( M \)
Now, we remove mass \( M \) from the system. The only mass left is \( m \). The new equilibrium position will now be determined by the weight of mass \( m \).

**Hint:** Consider how the removal of mass \( M \) changes the forces acting on the spring.

### Step 5: Determine the New Equilibrium Position
The new force due to gravity acting on the remaining mass \( m \) is \( mg \). The new equilibrium position \( x_1 \) can be found using:
\[ k x_1 = mg \]
Solving for \( x_1 \):
\[ x_1 = \frac{mg}{k} \]

**Hint:** Again, use the equilibrium condition for the new setup with only mass \( m \).

### Step 6: Calculate the Amplitude of Oscillations
The amplitude of oscillations after removing mass \( M \) is the difference between the initial elongation \( x_0 \) and the new elongation \( x_1 \):
\[ A = x_0 - x_1 \]
Substituting the values we found:
\[ A = \frac{(M + m)g}{k} - \frac{mg}{k} \]
This simplifies to:
\[ A = \frac{(M + m - m)g}{k} = \frac{Mg}{k} \]

**Hint:** Simplify the expression carefully to find the final amplitude.

### Final Answer
The amplitude of oscillations after removing mass \( M \) is:
\[ A = \frac{Mg}{k} \]