A metallic rod of length 1 m, young's modulus `3xx10^(11)Nm^(-3)` and density is clamped at its middle. Longitudinal stationary vibrations are produced in the rod with the total number of displacement nodes equal to 3. The frequency of vibrations is

Options

A. 30,000 Hz
B. 10,000 Hz
C. 3000 Hz
D. 1500 Hz

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript and apply the relevant physics concepts.

### Step 1: Understand the Setup
We have a metallic rod of length \( L = 1 \, \text{m} \), Young's modulus \( Y = 3 \times 10^{11} \, \text{N/m}^2 \), and density \( \rho = 7.5 \times 10^2 \, \text{kg/m}^3 \). The rod is clamped at its middle, producing longitudinal stationary vibrations with a total of 3 displacement nodes.

### Step 2: Identify the Nodes and Antinodes
When the rod is clamped at the middle, it creates a node at the center. With a total of 3 nodes, we can visualize the arrangement as follows:
- Node at the center (1st node)
- Antinode at both ends (2nd and 3rd nodes)

Thus, we have:
- Node 1: Center
- Antinode 1: Left end
- Antinode 2: Right end

### Step 3: Relate Length to Wavelength
The distance between nodes and antinodes is given by:
- Distance between a node and an antinode = \( \frac{\lambda}{4} \)
- Distance between two nodes = \( \frac{\lambda}{2} \)

Since there are 3 nodes, the total length of the rod can be expressed as:
\[ L = \frac{3\lambda}{2} \]
From this, we can solve for the wavelength \( \lambda \):
\[ \lambda = \frac{2L}{3} = \frac{2 \times 1 \, \text{m}}{3} = \frac{2}{3} \, \text{m} \]

### Step 4: Calculate the Velocity of Longitudinal Waves
The velocity \( v \) of longitudinal waves in the rod can be calculated using the formula:
\[ v = \sqrt{\frac{Y}{\rho}} \]
Substituting the values:
\[ v = \sqrt{\frac{3 \times 10^{11} \, \text{N/m}^2}{7.5 \times 10^2 \, \text{kg/m}^3}} = \sqrt{\frac{3 \times 10^{11}}{750}} \, \text{m/s} \]

### Step 5: Simplify the Velocity Calculation
Calculating the fraction:
\[ \frac{3 \times 10^{11}}{750} = 4 \times 10^8 \]
Thus,
\[ v = \sqrt{4 \times 10^8} = 2 \times 10^4 \, \text{m/s} \]

### Step 6: Calculate the Frequency
The frequency \( f \) can be calculated using the formula:
\[ f = \frac{v}{\lambda} \]
Substituting the values we found:
\[ f = \frac{2 \times 10^4 \, \text{m/s}}{\frac{2}{3} \, \text{m}} = 2 \times 10^4 \times \frac{3}{2} = 3 \times 10^4 \, \text{Hz} \]

### Final Answer
The frequency of the longitudinal vibrations in the rod is:
\[ f = 30000 \, \text{Hz} \]