A particle is attached to a vertical spring and is pulled down a distance 0.04m below its equilibrium position and is released from rest. The initial upward acceleration of the particle is `0.30 ms^(-2)`. The period of the oscillation is

Options

A. 4.08 s
B. 1.92 s
C. 3.90 s
D. 2.29 s

Explanation

To solve the problem, we need to find the period of oscillation of a particle attached to a vertical spring after being pulled down and released. Here’s a step-by-step solution:

### Step 1: Understand the given data
- The particle is pulled down a distance \( A = 0.04 \, \text{m} \) (this is the amplitude of oscillation).
- The initial upward acceleration \( a = 0.30 \, \text{m/s}^2 \).

### Step 2: Relate acceleration to angular frequency
In simple harmonic motion (SHM), the relationship between acceleration \( a \), angular frequency \( \omega \), and amplitude \( A \) is given by the formula:
\[ a = \omega^2 A \]
We can rearrange this to find \( \omega^2 \):
\[ \omega^2 = \frac{a}{A} \]

### Step 3: Substitute the values
Now, we can substitute the known values into the equation:
\[ \omega^2 = \frac{0.30 \, \text{m/s}^2}{0.04 \, \text{m}} = 7.5 \, \text{s}^{-2} \]

### Step 4: Calculate \( \omega \)
To find \( \omega \), we take the square root of \( \omega^2 \):
\[ \omega = \sqrt{7.5} \approx 2.74 \, \text{rad/s} \]

### Step 5: Find the period of oscillation
The period \( T \) of oscillation is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{2.74} \approx \frac{6.2832}{2.74} \approx 2.29 \, \text{s} \]

### Final Answer
The period of the oscillation is approximately \( 2.29 \, \text{s} \).

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