A block of mass `1kg` hangs without vibrations at the end of a spring with a force constant `1 N//m` attached to the ceilling of an elevator. The elevator is rising with an upward acceleration of `g//4`. The acceleration of the elevator suddenly ceases. What is the amplitude of the resulting oscillations?

Explanation

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

### Step 1: Understanding the Forces Acting on the Block
When the elevator is rising with an upward acceleration of \( \frac{g}{4} \), the effective force acting on the block due to gravity is modified by the pseudo force due to the elevator's acceleration.

The effective gravitational force \( F_g \) acting on the block while the elevator is accelerating is given by:
\[ F_g = mg + ma \]
where:
- \( m = 1 \, \text{kg} \) (mass of the block)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( a = \frac{g}{4} = \frac{10}{4} = 2.5 \, \text{m/s}^2 \) (upward acceleration of the elevator)

Substituting the values:
\[ F_g = 1 \times 10 + 1 \times 2.5 = 10 + 2.5 = 12.5 \, \text{N} \]

### Step 2: Finding the Spring Force at Equilibrium
At equilibrium, the spring force \( F_s \) is equal to the effective gravitational force:
\[ F_s = kx \]
where:
- \( k = 1 \, \text{N/m} \) (spring constant)
- \( x \) is the extension of the spring from its natural length.

Setting the forces equal gives:
\[ kx = 12.5 \]
Substituting \( k \):
\[ 1 \cdot x = 12.5 \implies x = 12.5 \, \text{m} \]

### Step 3: Analyzing the Situation When the Elevator Stops
When the elevator suddenly ceases to accelerate, the only force acting on the block is the gravitational force. The spring will now exert a force equal to the weight of the block:
\[ F_s = mg = 1 \times 10 = 10 \, \text{N} \]

### Step 4: Finding the Amplitude of Oscillation
The difference in forces when the elevator stops creates a net force that will cause the block to oscillate. The maximum extension (amplitude \( A \)) of the spring will be when the spring force equals the new effective gravitational force:
\[ F_s = kA \]
The net force when the elevator stops is:
\[ F_{\text{net}} = F_g - F_s = 12.5 - 10 = 2.5 \, \text{N} \]
Setting this equal to the spring force at maximum extension:
\[ kA = 2.5 \implies 1 \cdot A = 2.5 \implies A = 2.5 \, \text{m} \]

### Final Answer
The amplitude of the resulting oscillations is:
\[ \boxed{2.5 \, \text{m}} \]