Question

The range audible frequency for humans is $20Hzto20,000Hz.$ lf speed of sound in air is $336m/s$. The maximum length of a musical instrument, based on resonance pipe is $l_{max}=\frac{v}{x\times 20}{n}_{max}$. What is the value of x?

Answer 1

For open pipe frequency of over tones

$f=\frac{v}{2l}n\Rightarrow l=\frac{v}{2f}n\to \left(1\right)$

n= 1,2,3....

v= Velocity of sound

l= length of pipe

For closed pip

$f=\frac{v}{4l}(2x+1)\Rightarrow l=\frac{v}{4f}(2x+1)\to \left(1\right)$

For length to be maximum we have to use equation (1)

$l_{max}=\frac{v}{2f_{min}}{n}_{max}\Rightarrow l_{max}=\frac{v}{2\times 20}{n}_{max}$

$f_{min}=20HZ$ For minimum audible frequency

$x=2$