Question

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

• A. 2.5KHz
• B. 5 KHz
• C. 7.5 KHz
• D. 10 KHz

Answer 1

Answer: (B)

5 KHz

Answer 2

The correct answer is option (B): 5 KHz

Given:-
Y=$9.27×10^{10 }Pa$
ρ=$2.7×10^3 kg/m^3$
Length of rod L=60 cm=0.6 m

Fundamental frequency = $\frac1{2L} \sqrt{\frac{Y}{ρ}}$

= $\frac1{2(0.6)} \sqrt{\frac{9.27×10^{10 }}{2.7×10^3 }}$

==$4.9×10^3 Hz = 5kHz$