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 `5 m//s^(2)` when the acceleration suddenly ceases at time `t = 0` and the car moves upward with constant speed `(g = 10 m//s^(2))`
What is the angular frequencyof the block after the acceleration ceases ?

Options

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

Explanation

To solve the problem step by step, we need to find the angular frequency of the block after the elevator's upward acceleration ceases.

### Step 1: Identify the parameters
- Mass of the block, \( m = 2 \, \text{kg} \)
- Spring constant, \( k = 400 \, \text{N/m} \)

### Step 2: Understand the formula for angular frequency
The angular frequency \( \omega \) of a mass-spring system is given by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]

### Step 3: Substitute the values into the formula
Now, we can substitute the known values of \( k \) and \( m \) into the formula:
\[ \omega = \sqrt{\frac{400 \, \text{N/m}}{2 \, \text{kg}}} \]

### Step 4: Calculate the value
Calculating the fraction inside the square root:
\[ \frac{400}{2} = 200 \]
Now take the square root:
\[ \omega = \sqrt{200} \]

### Step 5: Simplify the square root
We can simplify \( \sqrt{200} \):
\[ \sqrt{200} = \sqrt{100 \times 2} = \sqrt{100} \times \sqrt{2} = 10\sqrt{2} \]

### Step 6: State the final answer
Thus, the angular frequency of the block after the acceleration ceases is:
\[ \omega = 10\sqrt{2} \, \text{rad/s} \]

### Conclusion
The correct answer is \( 10\sqrt{2} \, \text{rad/s} \).

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