A spring of force constant `200Nm^(-1)` has a block of mass 10 kg hanging at its one end and the other end of the spring is attached to the celling of an elevator. The elevator is rising upwards with an acceleration of `g/4` and the block is in equilibrium with respect to the elevator . when the acceleration of the elevator suddenly ceases , the block starts oscillating . What is the amplitude (in m) of these oscillations ?

Answer

To solve the problem step by step, we will analyze the forces acting on the block when the elevator is accelerating and when it suddenly stops.

### Step 1: Identify the forces acting on the block when the elevator is accelerating.

When the elevator is rising with an acceleration of \( g/4 \), the forces acting on the block are:
- The gravitational force acting downward, \( F_g = mg \)
- The spring force acting upward, \( F_s = kx \)
- A pseudo force acting downward due to the acceleration of the elevator, \( F_p = ma = m(g/4) \)

### Step 2: Write the equation for equilibrium in the accelerating frame.

In the accelerating frame of the elevator, the block is in equilibrium. Therefore, the sum of the forces acting on the block is zero:

\[ F_s = F_g + F_p \]

Substituting the forces, we have:

\[ kx_1 = mg + \frac{mg}{4} \]

### Step 3: Simplify the equation.

Combine the terms on the right side:

\[ kx_1 = mg + \frac{mg}{4} = mg \left(1 + \frac{1}{4}\right) = mg \cdot \frac{5}{4} \]

### Step 4: Solve for the extension \( x_1 \).

Now, we can solve for \( x_1 \):

\[ x_1 = \frac{5mg}{4k} \]

### Step 5: Identify the forces acting on the block when the elevator stops.

When the elevator suddenly stops, the only forces acting on the block are:
- The gravitational force \( F_g = mg \) acting downward
- The spring force \( F_s = kx_2 \) acting upward

### Step 6: Write the equation for equilibrium after the elevator stops.

At this point, the block is in equilibrium again, so we have:

\[ kx_2 = mg \]

### Step 7: Solve for the extension \( x_2 \).

Now we can solve for \( x_2 \):

\[ x_2 = \frac{mg}{k} \]

### Step 8: Calculate the amplitude of oscillation.

The amplitude \( A \) of the oscillation is the difference between the maximum extension \( x_1 \) and the equilibrium extension \( x_2 \):

\[ A = x_1 - x_2 \]

Substituting the expressions we derived:

\[ A = \frac{5mg}{4k} - \frac{mg}{k} \]

### Step 9: Factor out common terms.

Factoring out \( \frac{mg}{k} \):

\[ A = \frac{mg}{k} \left( \frac{5}{4} - 1 \right) = \frac{mg}{k} \left( \frac{5 - 4}{4} \right) = \frac{mg}{4k} \]

### Step 10: Substitute the values of \( m \), \( g \), and \( k \).

Given:
- \( m = 10 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \) (approximating \( g \))
- \( k = 200 \, \text{N/m} \)

Substituting these values:

\[ A = \frac{10 \cdot 10}{4 \cdot 200} = \frac{100}{800} = \frac{1}{8} \, \text{m} = 0.125 \, \text{m} \]

### Final Answer:

The amplitude of the oscillations is \( 0.125 \, \text{m} \).

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