Question

A metal rod of 50 cm length is clamped at its midpoint and is set to vibrations. The density of that metal is \( 2 \times 10^3 \, {kg/m}^3 \).
Young's modulus of that metal is \( 8 \times 10^8 \, {Nm}^2 \). The fundamental frequency of the vibration is 
 

• A. 2500 Hz
• B. 2 kHz
• C. 2.75 kHz
• D. 200 Hz

Hint:

The fundamental frequency of a vibrating rod depends on its length, material properties, and cross-sectional area.

Answer 1

The Correct Option is B

When a metal rod of length $L$ is clamped at its midpoint, the fundamental mode of vibration has a node at the center and antinodes at the ends. This means that the length of the rod corresponds to half a wavelength, so $\lambda = 2L$. The speed of a transverse wave in a solid is given by \[v = \sqrt{\frac{Y}{\rho}},\]where $Y$ is Young's modulus and $\rho$ is the density. The frequency $f$, wavelength $\lambda$, and speed $v$ of a wave are related by \[v = f \lambda.\]Hence, \[f = \frac{v}{\lambda} = \frac{1}{\lambda} \sqrt{\frac{Y}{\rho}} = \frac{1}{2L} \sqrt{\frac{Y}{\rho}}.\]We are given that $L = 50 \, {cm} = 0.5 \, {m}$, $\rho = 2 \times 10^3 \, {kg/m}^3$, and $Y = 8 \times 10^8 \, {Nm}^2$, so \[f = \frac{1}{2 \cdot 0.5} \sqrt{\frac{8 \times 10^8}{2 \times 10^3}} = \sqrt{4 \times 10^5} = 2 \times 10^2 = 2000 \, {Hz} = \boxed{2 \, {kHz}}.\] Final Answer: 2 kHz.