A block of mass `1kg` hangs without vibrating at the end of a spring whose force constant is `200(N)/(m)` and which is attached to the ceiling of an elevator. The elevator is rising with an upward acceleration of `(g)/(3)` when the acceleration suddenly ceases. The angular frequency of the block after the acceleration ceases is

Options

A. `13(rad)/(s)`
B. `14(rad)/(s)`
C. `15(rad)/(s)`
D. none of these

Explanation

To solve the problem, we need to find the angular frequency of a block attached to a spring after the elevator's upward acceleration ceases. Here’s a step-by-step solution:

### Step 1: Identify the parameters
We have:
- Mass of the block, \( m = 1 \, \text{kg} \)
- Spring constant, \( k = 200 \, \text{N/m} \)

### Step 2: Understand the motion
When the elevator is rising with an upward acceleration of \( \frac{g}{3} \), the block is in equilibrium and not vibrating. When the elevator suddenly stops, the block will start to oscillate in simple harmonic motion (SHM).

### Step 3: Recall 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 4: Substitute the values into the formula
Now, we can substitute the values of \( k \) and \( m \) into the formula:
\[ \omega = \sqrt{\frac{200 \, \text{N/m}}{1 \, \text{kg}}} \]

### Step 5: Calculate the angular frequency
Calculating the above expression:
\[ \omega = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \approx 14.14 \, \text{rad/s} \]

### Step 6: Conclusion
Thus, the angular frequency of the block after the elevator's acceleration ceases is approximately \( 14.14 \, \text{rad/s} \).