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Difficulty: Very hardResononace, Vibrating Strings, and Air Columns in Pipes

An open pipe of physical length 0.58 m0.58\text{ m} and a pipe closed at one end emit sound at the same frequency when both are vibrating in their first overtone mode. If the end correction at each open end is 0.01 m0.01\text{ m}, what is the physical length of the closed pipe?

  1. A
    0.29 m0.29\text{ m}
  2. B
    0.43 m0.43\text{ m}
  3. 0.44 m0.44\text{ m}Answer
  4. D
    0.45 m0.45\text{ m}

Answer

The physical length of the closed pipe is 0.44 m0.44\text{ m}.
The effective length of an open pipe with two open ends is L1,eff=0.58+2(0.01)=0.60 mL_{1,\text{eff}} = 0.58 + 2(0.01) = 0.60\text{ m}. The first overtone for an open pipe is its second harmonic (n=2n=2), giving f=v0.60f = \frac{v}{0.60}. For a pipe closed at one end, the first overtone is its third harmonic (m=3m=3), giving f=3v4L2,efff = \frac{3v}{4L_{2,\text{eff}}}. Equating frequencies gives L2,eff=0.45 mL_{2,\text{eff}} = 0.45\text{ m}. Subtracting the single end correction (e=0.01 me = 0.01\text{ m}) yields the physical length 0.44 m0.44\text{ m}.

Step-by-Step Solution

1
Calculate the effective length and first overtone frequency of the open pipe.
L1,eff=L1+2e=0.58 m+2(0.01 m)=0.60 mL_{1,\text{eff}} = L_1 + 2e = 0.58\text{ m} + 2(0.01\text{ m}) = 0.60\text{ m}, so fopen,1st overtone=2v2L1,eff=v0.60f_{\text{open,1st overtone}} = \frac{2v}{2L_{1,\text{eff}}} = \frac{v}{0.60}.
An open pipe has two open ends, so end correction is applied at both ends (2e2e). Its first overtone corresponds to the second harmonic (n=2n=2).
2
Express the first overtone frequency of the closed pipe in terms of its effective length.
fclosed,1st overtone=3v4L2,efff_{\text{closed,1st overtone}} = \frac{3v}{4L_{2,\text{eff}}}.
A pipe closed at one end produces only odd harmonics (m=1,3,5,…m=1, 3, 5, \dots). The first overtone corresponds to the third harmonic (m=3m=3).
3
Equate the two frequencies and solve for the effective length of the closed pipe.
v0.60=3v4L2,eff  ⟹  4L2,eff=1.80 m  ⟹  L2,eff=0.45 m\frac{v}{0.60} = \frac{3v}{4L_{2,\text{eff}}} \implies 4L_{2,\text{eff}} = 1.80\text{ m} \implies L_{2,\text{eff}} = 0.45\text{ m}.
Both pipes emit sound at the same frequency in their first overtone modes.
4
Calculate the physical length of the closed pipe.
L2=L2,eff−e=0.45 m−0.01 m=0.44 mL_2 = L_{2,\text{eff}} - e = 0.45\text{ m} - 0.01\text{ m} = 0.44\text{ m}.
A closed pipe has only one open end, so its effective length is L2+eL_2 + e.

Key Concept

Standing Waves and End Correction in Open and Closed Organ Pipes
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Topics

Resononace, Vibrating Strings, and Air Columns in Pipes

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