A granite rod of 60 cm length is clamped at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In solids,the velocity of a longitudinal wave is given by $v = \sqrt{\frac{Y}{ho}}$.Substituting the given values: $v = \sqrt{\frac{9.27 	imes

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) In solids,the velocity of a longitudinal wave is given by $v = \sqrt{\frac{Y}{ho}}$.Substituting the given values: $v = \sqrt{\frac{9.27 	imes 10^{10}}{2.7 	imes 10^3}} = \sqrt{3.433 	imes 10^7} \approx 5859 \ m/s$.Since the rod is clamped at its middle,the middle point acts as a node $(N)$ and the ends act as antinodes $(A)$. For the fundamental mode,the length of the rod $L$ corresponds to half a wavelength,so $L = \frac{\lambda}{2}$,which implies $\lambda = 2L$.Given $L = 60 \ cm = 0.6 \ m$,we have $\lambda = 2 	imes 0.6 = 1.2 \ m$.The fundamental frequency is $f = \frac{v}{\lambda} = \frac{5859}{1.2} \approx 4882.5 \ Hz$.Converting to $kHz$,$f \approx 4.88 \ kHz$,which is approximately $5 \ kHz$.

$A$ 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 modulus is $9.27 \times 10^{10} \ Pa$. What will be the fundamental frequency of the longitudinal vibrations in $kHz$?

Similar Questions

$A$ stone is dropped into a well that is $19.6\,m$ deep. If the echo is heard after $2.06\,s$ of dropping the stone,calculate the velocity of sound in $m/s$.

When an aeroplane attains a speed higher than the velocity of sound in air,a loud bang is heard. This is because

At constant pressure,out of $H_2$ and $O_2$ gas,in which gas is the speed of sound greater?

Why is the speed of sound higher in solids and liquids as compared to gases?

Vedclass Test Series

Exam Paper Generator

Online Exam Module