Question

A granite rod of $60\, cm$ length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $2.7 \times 10^3 \, kg/m^3$ and its Young's modules is $9.27 \times 10^{10}\, Pa$. What will be the fundamental frequency of the longitudinal vibrations ?

• A. 5 kHz
• B. 2.5 kHz
• C. 10 kHz
• D. 7.5 kHz

Answer 1

The Correct Option is A

To determine the fundamental frequency of the longitudinal vibrations of the granite rod, we will use the formula for the fundamental frequency of a rod clamped at its center, vibrating in its fundamental mode:

f = \frac{v}{2L}

Where:

• f is the fundamental frequency.
• v is the speed of sound in the material.
• L is the length of the rod.

The speed of sound in a material is given by:

v = \sqrt{\frac{Y}{\rho}}

Where:

• Y is the Young's modulus of the material.
• \rho is the density of the material.

Let's plug in the known values:

• Y = 9.27 \times 10^{10} \, \text{Pa}
• \rho = 2.7 \times 10^3 \, \text{kg/m}^3
• L = 60 \, \text{cm} = 0.6 \, \text{m}

First, calculate the speed of sound:

v = \sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^3}}

Calculating the above gives:

v = \sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^3}} \approx \sqrt{3.433 \times 10^7} \approx 5858.6 \, \text{m/s}

Now, calculate the fundamental frequency:

f = \frac{5858.6}{2 \times 0.6}

Calculating the above gives:

f = \frac{5858.6}{1.2} \approx 4882.17 \, \text{Hz} \approx 5 \, \text{kHz}

Thus, the fundamental frequency of the longitudinal vibration of the granite rod is approximately 5 kHz. Therefore, the correct option is 5 kHz.