A rod 100 cm in length and of material of density `8 xx 10^(3)" kg/m"^3` is clamped at the centre and attached to a Kundt's tube containing air. The distance between the first and the tenth node when the rod is stroked longitudinally is 135 cm. If the velocity of sound in air is 330 m/s determine the Young's modulus of the material of the rod.

Answer

To determine the Young's modulus of the material of the rod, we can follow these steps:

### Step 1: Understand the setup
The rod is clamped at the center, creating nodes at the center and antinodes at the ends. The distance between the first and the tenth node is given as 135 cm.

### Step 2: Calculate the wavelength
The distance between nodes in a standing wave is half the wavelength (λ/2). Since the distance between the first and the tenth node is 9 half-wavelengths (from node 1 to node 10), we can express this as:
\[ \text{Distance} = 9 \times \frac{\lambda}{2} \]
Given that this distance is 135 cm (or 1.35 m), we can set up the equation:
\[ 1.35 = 9 \times \frac{\lambda}{2} \]
Solving for λ:
\[ \lambda = \frac{1.35 \times 2}{9} = 0.3 \, \text{m} \]

### Step 3: Calculate the frequency
Using the speed of sound in air (v = 330 m/s) and the wavelength (λ = 0.3 m), we can find the frequency (f) using the formula:
\[ v = f \lambda \]
Rearranging gives:
\[ f = \frac{v}{\lambda} = \frac{330}{0.3} = 1100 \, \text{Hz} \]

### Step 4: Use the formula for Young's modulus
The formula for Young's modulus (Y) in terms of frequency, density (ρ), and wavelength is:
\[ Y = \frac{(2\pi f)^2 \cdot \rho \cdot \lambda^2}{4} \]
Substituting the values:
- \( f = 1100 \, \text{Hz} \)
- \( \rho = 8 \times 10^3 \, \text{kg/m}^3 \)
- \( \lambda = 0.3 \, \text{m} \)

### Step 5: Calculate Young's modulus
Substituting into the formula:
\[ Y = \frac{(2\pi \times 1100)^2 \cdot (8 \times 10^3) \cdot (0.3)^2}{4} \]
Calculating \( (2\pi \times 1100)^2 \):
\[ (2\pi \times 1100)^2 \approx (6911.5)^2 \approx 47851.4 \, \text{(approximately)} \]
Now substituting back:
\[ Y = \frac{47851.4 \cdot 8 \times 10^3 \cdot 0.09}{4} \]
Calculating:
\[ Y = \frac{47851.4 \cdot 72000}{4} \approx \frac{3440000000}{4} \approx 860000000 \, \text{N/m}^2 \]
Thus, the Young's modulus \( Y \approx 8.6 \times 10^8 \, \text{N/m}^2 \).

### Final Answer
The Young's modulus of the material of the rod is approximately \( 8.6 \times 10^8 \, \text{N/m}^2 \).

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