Question

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

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

Answer 1

The correct option is C

$5kHz$

Explanation for the correct option:

Step 1: Given data:

Length of a granite rod, $$\begin{gathered} L=60cm \\ =0.6m \end{gathered}$$

density of granite is $\rho =2.7x{10}^{3}kg/m^3$$\rho =2.7x{10}^{3}kg/{m}^{3}$

Young’s modulus is $Y=9.27x{10}^{10}Pa$

Step 2: Finding the fundamental frequency of the longitudinal vibrations:

Velocity of wave, $v=\sqrt{\frac{Y}{\rho }}$………..$\left(1\right)$​​

$$\begin{gathered} =\sqrt{\frac{9.27\times {10}^{10}}{2.7\times {10}^3}} \\ v=5.85x{10}^{3}m/s \end{gathered}$$

We know, $L=\frac{\lambda }{2}$ [Where, $\lambda$is wavelength]

$\Rightarrow \lambda =2L$………..$\left(2\right)$

Frequency, $\begin{array}{l}f=\frac{v}{\lambda }\end{array}$

$$\begin{gathered} \begin{array}{l}=\frac{1}{2L}\sqrt{\frac{Y}{p}}\end{array} \\ =\frac{1}{2x0.6}x5.85x{10}^3 \\ =4.88x{10}^{3}Hz\simeq 5kHz \end{gathered}$$

Hence, option $\left(C\right)$ is correct.