Question

An aluminium rod having length $100\text{cm}$ is clamped at its middle point and set into longitudinal vibrations. Let the rod vibrate in its fundamental mode. The density of aluminium is $2600{\text{kg/m}}^3$ and its Young's modulus is $7.8\times {10}^{10}{\text{N/m}}^2.$ The frequency of the sound produced is

• A. $1250\text{Hz}$
• B. $2738\text{Hz}$
• C. $2350\text{Hz}$
• D. $1685\text{Hz}$

Answer 1

The correct option is B $2738\text{Hz}$

The frequency of sound produced will be equal to the fundamental frequency of oscillation of the rod.
Given:
Length of the aluminium rod, $l=100\text{cm}=1\text{m}$
Young's modulus of Aluminium rod is
$Y=7.8\times {10}^{10}{\text{N/m}}^2$
Density of the aluminium, $\rho =2600{\text{kg/m}}^3$
We know that velocity of sound is
$v=\sqrt{\frac{Y}{\rho }}$
$\Rightarrow v=\sqrt{\frac{7.8\times {10}^{10}}{2600}}$
$\Rightarrow v\approx 5477\text{m/s}$
Rod is clamped at the middle, so, the middle point will act as displacement node and pressure antinode. Also, the ends of the rod will act as the pressure node.
We know that the distance between two successive nodes in a fundamental mode is $\lambda /2$.
$\Rightarrow \lambda /2=l$
$\Rightarrow \lambda =2l=2\times 1=2\text{m}$
Now,
$v=\lambda f$
$\Rightarrow f=\frac{v}{\lambda }$
$\Rightarrow f=\frac{5477}{2}\approx 2738\text{Hz}$