An open pipe has a length of 0.75 m. What would be the length of a closed organ pipe whose third harmonic(n=3) is the same as the fundamental frequency of the open pipe?

1 Answer
Sep 14, 2015

Answer:

#3.4"m"#

Explanation:

For an open pipe the 1st harmonic looks like this:

www.studyphysics.ca

This refers to half a wavelength so:

#L_1=lambda/2#

#lambda=2L_1#

Since #v=flambda#:

#f=v/lambda=(v)/(2L_1)##" "##color(red)((1))#

For the closed pipe the 3rd harmonic looks like this:

www.studyphysics.ca

In this case:

#L_2=3/4lambda#

#lambda=(4L_2)/(3)#

So #v=f.(4L_2)/(3)#

So #f=(3v)/(4L_2)# #" "##color(red)((2))#

These frequencies are equal so putting #color(red)((1))# equal to #color(red)((2))rArr#

#(v)/(2L_1)=(3v)/(4L_2)#

#(3)/(4L_2)=(1)/(2L_1)#

#L_2=(6L_1)/(4)#

#L_2=(3xx6xx0.75)/(4)#

#L_2=3.4"m"#