Question

A granite rod of $60cm$ length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $2.7 \times {10^3}kg/{m^3}$ and its Young’s modulus is $9.27 \times {10^{10}}Pa.$ What will be the fundamental frequency of the longitudinal vibrations?
A. $10kHz$
B. $7.5kHz$
C. $5kHz$
D. $2.5kHz$

Answer 1

It is a direct formula based question. Use the formula of fundamental frequency and velocity of sound to find the answer.
$V = \sqrt {\dfrac{Y}{\ell }}$
$\upsilon = \dfrac{V}{\lambda }$

Complete step by step answer:
It is given that a granite red is clamped at tee middle

Also, Young’s modulus, $Y = 9.27 \times {10^{10}}$
The density of granite, $\ell = 2.7 \times {10^3}kg/{m^3}$
Length of granite rod$,L = 60cm$
$\Rightarrow L = 0.6m$ ($1m = 100cm$)
Let $V$be the velocity of sound in granite.
We know that
$\upsilon = \sqrt {\dfrac{Y}{\ell }}$
Substituting the values, we get
$V = \sqrt {\dfrac{{9.27 \times {{10}^{10}}}}{{2.7 \times {{10}^3}}}}$
By rearranging we get,
$V = \sqrt {\dfrac{{927 \times {{10}^{ - 2}} \times {{10}^{10}}}}{{27 \times {{10}^{ - 1}} \times {{10}^3}}}}$
$= \sqrt {\dfrac{{927 \times {{10}^{10 - 2}}}}{{27 \times {{10}^{3 - 1}}}}}$ $\left( {\because {a^m}{a^n} = {a^{m + n}}} \right)$
$= \sqrt {\dfrac{{927 \times {{10}^8}}}{{27 \times {{10}^2}}}}$
$= \sqrt {34.33} \times {10^{8 - 2}}$ $\left( {\because \dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right)$
$= \sqrt {34.33 \times {{10}^6}}$
$\Rightarrow V \approx 5.9 \times {10^3}m/s.$
Substituting this value in equation (1) we get
Fundamental frequency$= \dfrac{{5.9 \times {{10}^3}}}{{2 \times 0.6}}$
$= \dfrac{{5.9 \times {{10}^3}}}{{1.2}}$
$= \dfrac{{59}}{{12}} \times {10^3}$
$= 4.91 \times {10^3}Hz$
$\upsilon = 4.91 \times {10^3}Hz$
$\upsilon \approx 5kHz$ $(\because 1kHz = {10^3}Hz)$

So, the correct answer is “Option C”.

Note:
Sometimes it becomes difficult to find the exact values in the numerical of physics. In such cases we can check for the closest possible value, else we did in this question. Do not waste your time in trying to find the exact answers of such questions. We can use log table complex calculations.