A metallic rod of length 1 m, young's modulus `3xx10^(11)Nm^(-3)` and density is `7.5xx10^2 (kg)/m^3`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. (a)30,000 Hz
B. (b)10,000 Hz
C. (c)3000 Hz
D. (d)1500 Hz

Answer

To find the frequency of longitudinal vibrations in the metallic rod, we can follow these steps:

### Step 1: Understand the Setup
We have a metallic rod of length \( L = 1 \, \text{m} \) that is clamped at its middle, producing longitudinal stationary vibrations. The total number of displacement nodes is given as 3.

### Step 2: Determine the Node Configuration
Since the rod is clamped at the middle, it will have a node at the center. With 3 nodes in total, the configuration will look like this:
- Node at \( x = 0 \) (clamped end)
- Node at \( x = \frac{L}{3} \)
- Node at \( x = \frac{2L}{3} \)
- Node at \( x = L \) (free end)

### Step 3: Relate Length to Wavelength
For a rod with 3 nodes, the distance between the nodes corresponds to a fraction of the wavelength. The distance from the first node to the second node is \( \frac{\lambda}{2} \) and from the second node to the third node is also \( \frac{\lambda}{2} \). The total length of the rod can be expressed as:
\[ L = 3 \frac{\lambda}{4} \]
This means:
\[ \frac{3\lambda}{4} = 1 \, \text{m} \implies \lambda = \frac{4}{3} \, \text{m} \]

### Step 4: Calculate the Velocity of Longitudinal Waves
The velocity \( v \) of longitudinal waves in a material is given by the formula:
\[ v = \sqrt{\frac{Y}{\rho}} \]
where \( Y \) is the Young's modulus and \( \rho \) is the density of the material.

Given:
- \( Y = 3 \times 10^{11} \, \text{N/m}^2 \)
- \( \rho = 7.5 \times 10^2 \, \text{kg/m}^3 \)

Substituting these values:
\[ v = \sqrt{\frac{3 \times 10^{11}}{7.5 \times 10^2}} = \sqrt{4 \times 10^{8}} = 2 \times 10^4 \, \text{m/s} \]

### Step 5: Calculate the Frequency
The frequency \( f \) can be calculated using the relationship:
\[ f = \frac{v}{\lambda} \]
Substituting the values:
\[ f = \frac{2 \times 10^4}{\frac{4}{3}} = 2 \times 10^4 \times \frac{3}{4} = 1.5 \times 10^4 \, \text{Hz} \]

### Final Answer
The frequency of vibrations in the rod is:
\[ f = 1.5 \times 10^4 \, \text{Hz} \]