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 involving the two masses and the spring, and then derive the amplitude of oscillation after one mass is removed.

### Step 1: Understand the Initial Equilibrium Condition
When both masses \( M \) and \( m \) are suspended together by the spring, the spring is stretched to a new equilibrium position due to the combined weight of the two masses. The force balance at equilibrium can be expressed as:

\[ k x_0 = (M + m) g \]

Where:
- \( k \) is the spring constant,
- \( x_0 \) is the initial extension of the spring,
- \( g \) is the acceleration due to gravity.

### Step 2: Remove Mass \( M \)
When mass \( M \) is removed, the system will adjust to a new equilibrium position. The only mass remaining is \( m \). The new equilibrium condition can be expressed as:

\[ k x_0' = m g \]

Where \( x_0' \) is the new extension of the spring after mass \( M \) is removed.

### Step 3: Solve for the New Extension \( x_0' \)
From the equation for the new equilibrium, we can solve for \( x_0' \):

\[ x_0' = \frac{m g}{k} \]

### Step 4: Find the Original Extension \( x_0 \)
From the initial equilibrium condition, we can express \( x_0 \) as:

\[ x_0 = \frac{(M + m) g}{k} \]

### Step 5: Calculate the Change in Extension
The change in extension of the spring when mass \( M \) is removed is given by:

\[ \Delta x = x_0 - x_0' = \frac{(M + m) g}{k} - \frac{m g}{k} \]

### Step 6: Simplify the Expression for Change in Extension
Now, simplifying the expression for \( \Delta x \):

\[ \Delta x = \frac{(M + m) g - m g}{k} = \frac{M g}{k} \]

### Step 7: Determine the Amplitude of Oscillation
The amplitude of oscillation after mass \( M \) is removed is equal to the change in extension \( \Delta x \):

\[ A = \Delta x = \frac{M g}{k} \]

Thus, the amplitude of oscillation after removing mass \( M \) is:

\[ \boxed{\frac{M g}{k}} \]