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Wize University Physics Textbook (Master) > Waves: Mechanical

Standing Waves in Air Pipes

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Standing Waves in Air Pipes


Standing waves could be formed in air pipes and the patterns and relation between the length of the pipe and resonant frequencies depends on the boundary conditions at the ends of the pipe


  • Closed ends of the air pipes form nodes

  • Open ends of the air pipes form anti-nodes



Two Ends Closed

  • The boundary conditions for this case give zero displacement at the endpoints. So, we have nodes at each closed end.
  • The simplest pattern possible to satisfy the boundary conditions, corresponding to the longest wavelength, has L=λ2L=\dfrac{\lambda}{2}











The wavelengths are:
 λn=2Ln \boxed{ \ \lambda_n=\dfrac{2L}{n} \ }


and the corresponding frequencies are:

 fn=nv2L \boxed{ \ f_n=\dfrac{nv}{2L}\ }
for n=1,2,3,…n=1,2,3,\ldots




Wize Concept
The configuration of the waves inside a pipe with two closed ends is the same as for a string with both ends fixed, and the equations are exactly the same.

Two Ends Open

  • The boundary conditions now give the highest displacement at the endpoints. So, we have antinodes at both ends.
  • The simplest pattern possible to satisfy the boundary conditions, corresponding to the longest wavelength, has L=λ2L=\dfrac{\lambda}{2}















The wavelengths are:
 λn=2Ln \boxed{ \ \lambda_n=\dfrac{2L}{n} \ }


and the corresponding frequencies are:

 fn=nv2L \boxed{ \ f_n=\dfrac{nv}{2L}\ }
for n=1,2,3,…n=1,2,3,\ldots




Wize Concept
Even though the configuration of the waves inside the pipe is different for two open ends vs. two closed ends, the number of half wavelengths is the same for both cases, so the equations are actually the same.

One End Open & One End Closed

  • In this case, we need to have a node at the closed end and an antinode at the open end
  • The simplest pattern possible to satisfy the boundary conditions, corresponding to the longest wavelength, has L=λ4L=\dfrac{\lambda}{4}















The wavelengths are:
 λn=4Ln \boxed{ \ \lambda_n=\dfrac{4L}{n} \ }
for n=1,3,5,…(odd numbers only)n=1,3,5,\ldots (odd \ numbers \ only)



and the corresponding frequencies are:

 fn=nv4L \boxed{ \ f_n=\dfrac{nv}{4L}\ }

for n=1,3,5,…(odd numbers only)n=1,3,5,\ldots (odd \ numbers \ only)


PAGE BREAK


Watch Out!
This is the only case that only has the odd harmonics. The even harmonics are missing. We can only have n=oddn= odd.






Exam Tip
This is the only case that uses a "4\bf4" instead of a "2\bf2" in all formulas.




Example: Standing Waves in Air Pipes


The third harmonic in a musical air pipe with one end open and one end closed is equal to 12501250 Hz.
(Use the speed of the sound value of 334334 m/s)

a) What is the length of the air pipe?
b) Draw the pattern corresponding to the third overtone . What harmonic is this?


Part a)

For a pipe open at one end and closed at the other, we count the quarter wavelengths: the third harmonic means n=3n=3, and the pattern has 33 quarter wavelengths:


Use the frequency formula to solve for the length:

fn=nv4L   →   L=nv4fnf_n=\dfrac{nv}{4L} \ \ \ \to \ \ \ L=\dfrac{nv}{4f_n}

=3⋅3344⋅1250=\dfrac{3\cdot334}{4\cdot1250}

=0.2004=0.2004 (m)


Part b)

3rd overtone = 3 tones over the fundamental
= 3 half wavelengths (or loops) over the fundamental
= 7 quarter wavelengths in total
= 7th harmonic

OR: n=1+3⋅2=7n=1+3\cdot2=7



Practice: Standing Waves in Air Pipes


A horn of 8585 cm length is open at one end and closed at the other end.
(Use the speed of the sound value of 334334 m/s)

a) What is the fundamental frequency of the horn?
b) What are the frequencies for the next two resonances?
c) By how much do you need to change the length in order to increase the fundamental frequency by 2020 Hz?

Part a)