A particle of mass 3 kg , attached to a spring with force constant `48Nm^(-1) `execute simple harmonic motion on a frictionless horizontal surface. The time period of oscillation of the particle, is seconds , is

Options

A. (a)`pi/4`
B. (b)`pi/2`
C. (c)`2pi`
D. (d)`8pi`

Answer

To solve the problem of finding the time period of oscillation for a particle attached to a spring, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the particle, \( m = 3 \, \text{kg} \)
- Spring constant, \( k = 48 \, \text{N/m} \)

2. **Use the Formula for Time Period:**
The time period \( T \) of a mass-spring system executing simple harmonic motion is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

3. **Substitute the Given Values into the Formula:**
Plugging in the values of \( m \) and \( k \):
\[ T = 2\pi \sqrt{\frac{3 \, \text{kg}}{48 \, \text{N/m}}} \]

4. **Simplify the Fraction Inside the Square Root:**
First, calculate \( \frac{3}{48} \):
\[ \frac{3}{48} = \frac{1}{16} \]
Therefore, we can rewrite the equation as:
\[ T = 2\pi \sqrt{\frac{1}{16}} \]

5. **Calculate the Square Root:**
The square root of \( \frac{1}{16} \) is:
\[ \sqrt{\frac{1}{16}} = \frac{1}{4} \]

6. **Substitute Back into the Time Period Formula:**
Now substituting back:
\[ T = 2\pi \cdot \frac{1}{4} \]

7. **Simplify the Expression:**
This simplifies to:
\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{seconds} \]

### Final Answer:
The time period of oscillation of the particle is \( \frac{\pi}{2} \) seconds.