A particle of mass 0.1 kg is held between two rigid supports by two springs of force constant `8 N m^(-1) and 2 N m^(-1)`. If the particle is displaced along the direction of length of the springs, its frequency of vibration is

Options

A. `5/pi`Hz
B. `8/pi` Hz
C. `2/pi`Hz
D. `1/pi` Hz

Answer

To find the frequency of vibration of the particle held between two springs, we can follow these steps:

### Step 1: Identify the given data
- Mass of the particle, \( m = 0.1 \, \text{kg} \)
- Spring constant of the first spring, \( k_1 = 8 \, \text{N/m} \)
- Spring constant of the second spring, \( k_2 = 2 \, \text{N/m} \)

### Step 2: Determine the equivalent spring constant
Since the two springs are in parallel, the equivalent spring constant \( k \) can be calculated as:
\[ k = k_1 + k_2 \]
Substituting the values:
\[ k = 8 \, \text{N/m} + 2 \, \text{N/m} = 10 \, \text{N/m} \]

### Step 3: Use the formula for frequency of oscillation
The frequency of oscillation \( f \) for a mass-spring system is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

### Step 4: Substitute the values into the formula
Now substituting the equivalent spring constant and the mass into the frequency formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{10 \, \text{N/m}}{0.1 \, \text{kg}}} \]

### Step 5: Simplify the expression
Calculating the value inside the square root:
\[ \frac{10}{0.1} = 100 \]
So,
\[ f = \frac{1}{2\pi} \sqrt{100} \]
\[ f = \frac{1}{2\pi} \times 10 \]
\[ f = \frac{10}{2\pi} = \frac{5}{\pi} \, \text{Hz} \]

### Final Answer
The frequency of vibration of the particle is:
\[ f = \frac{5}{\pi} \, \text{Hz} \]