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 can follow these instructions:

### Step 1: Understand the Forces Acting on the Block
The block of mass \( m = 1 \, \text{kg} \) is hanging from a spring with a spring constant \( k = 1 \, \text{N/m} \). The elevator is accelerating upward with an acceleration of \( \frac{g}{4} \), where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 2: Calculate the Effective Force on the Block
When the elevator is accelerating upward, the effective gravitational force acting on the block is increased due to the upward acceleration. The effective force can be calculated as:
\[ F_{\text{effective}} = m(g + a) \]
where \( a = \frac{g}{4} \). Thus,
\[ F_{\text{effective}} = 1 \, \text{kg} \left(10 \, \text{m/s}^2 + \frac{10 \, \text{m/s}^2}{4}\right) = 1 \, \text{kg} \left(10 + 2.5\right) = 1 \, \text{kg} \cdot 12.5 \, \text{m/s}^2 = 12.5 \, \text{N} \]

### Step 3: Relate the Force to the Spring Constant and Amplitude
When the elevator suddenly stops, the block will oscillate. The force exerted by the spring when it is stretched by an amplitude \( A \) is given by Hooke's Law:
\[ F = kA \]
Setting the effective force equal to the spring force gives:
\[ kA = 12.5 \, \text{N} \]

### Step 4: Solve for Amplitude \( A \)
Substituting the value of the spring constant \( k = 1 \, \text{N/m} \):
\[ 1 \, \text{N/m} \cdot A = 12.5 \, \text{N} \]
Thus,
\[ A = 12.5 \, \text{m} \]

### Step 5: Conclusion
The amplitude of the resulting oscillations when the elevator ceases to accelerate is \( A = 12.5 \, \text{m} \).