A particle is attached to a vertical spring and is pulled down a distance `4 cm` below its equilibrium and is released from rest. The initial upward acceleration is `0.5ms^(-2)`. The angular frequency of oscillation is

Options

A. `3.53 rad//s`
B. `0.28 rad//s`
C. `1.25 rad//s`
D. `0.08 rad//s`

Explanation

To find the angular frequency of oscillation for the particle attached to a vertical spring, we can follow these steps:

### Step 1: Understand the given data
- The particle is pulled down a distance \( x = 4 \, \text{cm} \) below its equilibrium position.
- The initial upward acceleration \( a = 0.5 \, \text{m/s}^2 \).

### Step 2: Convert the distance from centimeters to meters
Since the standard unit for distance in physics is meters, we convert \( 4 \, \text{cm} \) to meters:
\[ x = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m} = 0.04 \, \text{m} \]

### Step 3: Use the formula for angular frequency
The angular frequency \( \omega \) can be calculated using the formula:
\[ \omega = \sqrt{\frac{a}{x}} \]
where:
- \( a \) is the acceleration,
- \( x \) is the displacement from the equilibrium position.

### Step 4: Substitute the values into the formula
Now, substituting the values of \( a \) and \( x \):
\[ \omega = \sqrt{\frac{0.5 \, \text{m/s}^2}{0.04 \, \text{m}}} \]

### Step 5: Calculate the value
Calculating the fraction inside the square root:
\[ \frac{0.5}{0.04} = 12.5 \]
Now take the square root:
\[ \omega = \sqrt{12.5} \approx 3.54 \, \text{rad/s} \]

### Step 6: Final answer
Thus, the angular frequency of oscillation is approximately:
\[ \omega \approx 3.54 \, \text{rad/s} \]

### Summary of the solution
The angular frequency of oscillation for the particle attached to the vertical spring is approximately \( 3.54 \, \text{rad/s} \).

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