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. `pi/4`
B. `pi/2`
C. `2pi`
D. `8pi`

Answer

To find the time period of oscillation for a particle attached to a spring, we can use the formula for the time period \( T \) of a simple harmonic oscillator:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

where:
- \( T \) is the time period,
- \( m \) is the mass of the particle,
- \( k \) is the spring constant.

### Step 1: Identify the values given in the problem
- Mass of the particle, \( m = 3 \, \text{kg} \)
- Spring constant, \( k = 48 \, \text{N/m} \)

### Step 2: Substitute the values into the formula
Now we will substitute the values of \( m \) and \( k \) into the formula:

\[ T = 2\pi \sqrt{\frac{3}{48}} \]

### Step 3: Simplify the fraction inside the square root
First, simplify \( \frac{3}{48} \):

\[ \frac{3}{48} = \frac{1}{16} \]

### Step 4: Calculate the square root
Now, take the square root of \( \frac{1}{16} \):

\[ \sqrt{\frac{1}{16}} = \frac{1}{4} \]

### Step 5: Substitute back into the formula
Now substitute back into the formula for \( T \):

\[ T = 2\pi \cdot \frac{1}{4} \]

### Step 6: Simplify the expression
Now simplify:

\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \]

### Step 7: Final answer
Thus, the time period of oscillation of the particle is:

\[ T = \frac{\pi}{2} \, \text{seconds} \]