Passage X) A 2kg block hangs without vibrating at the bottom end of a spring with a force constant of 400 N/m. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of `5m/s^(2)` when the acceleration suddenly ceases at t=0 and the car moves upward with constant speed. (g=10m/`s^(2)`)
What is the angular frequency of oscillation of the block after the acceleration ceases?

Options

A. `(a)10sqrt(2)`rad/s
B. `(b)20 rad/s`
C. `(c)20sqrt(2)`rad/s
D. `(d)32 rad/s`

Answer

To find the angular frequency of oscillation of the block after the acceleration of the elevator ceases, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Mass of the block, \( m = 2 \, \text{kg} \)
- Spring constant, \( k = 400 \, \text{N/m} \)

2. **Understand the Concept**:
- The angular frequency \( \omega \) of a mass-spring system in simple harmonic motion is given by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
- This formula applies regardless of the motion of the elevator, as long as the spring is not stretched or compressed beyond its equilibrium position.

3. **Substitute the Values into the Formula**:
- Plugging in the values of \( k \) and \( m \):
\[ \omega = \sqrt{\frac{400 \, \text{N/m}}{2 \, \text{kg}}} \]

4. **Calculate the Angular Frequency**:
- Simplifying the expression:
\[ \omega = \sqrt{200} \, \text{rad/s} \]

5. **Express the Result**:
- The square root of 200 can be simplified further:
\[ \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \, \text{rad/s} \]

6. **Final Answer**:
- Therefore, the angular frequency of oscillation of the block after the acceleration ceases is:
\[ \omega = 10\sqrt{2} \, \text{rad/s} \]