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Worked Examples · Example 6.2

Q.A closed pipe and an open pipe have the same length. Show that no mode of the closed pipe has the same wavelength as any mode of the open pipe.

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Closed λᶜₚ = 4L/(2p+1); open λᵒₘ = 2L/(m+1). Setting them equal needs 2(m+1) = 2p+1, i.e. even = odd, which is impossible.

For a closed pipe the allowed frequencies are nᶜₚ = (2p+1)nᶜ with nᶜ = v/4L, so the wavelengths are

λpc=4L2p+1(p=0,1,2,… )\lambda^c_p = \frac{4L}{2p+1}\quad(p = 0, 1, 2, \dots)

For an open pipe nᵒₘ = (m+1)nᵒ with nᵒ = v/2L, so

λmo=2Lm+1(m=0,1,2,… )\lambda^o_m = \frac{2L}{m+1}\quad(m = 0, 1, 2, \dots)

If a closed mode had the same wavelength as an open mode,

4L2p+1=2Lm+1⇒2(m+1)=2p+1\frac{4L}{2p+1} = \frac{2L}{m+1} \Rightarrow 2(m+1) = 2p+1 …

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