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Fundamental frequency

  1. Jun 7, 2005 #1
    A certain pipe produces a fundamental frequency f in air.

    If the pipe is filled with helium at the same temperature, what fundamental frequency does it produce? (Take the molar mass of air to be M_air, and the molar mass of helium to be M_He). The ratio γ of heat capacities for air (7/5) and for Helium (5/3).

    I'm not sure how to approach this question.

    Any help would be great.

  2. jcsd
  3. Jun 7, 2005 #2


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    First, you have to find the Length of the pipe. Remember,
    [tex] c=f \lambda [/tex]

    and [tex] c= \sqrt\frac{\gamma P}{e} [/tex]

    also, [tex] PV=nRT [/tex]
    => [tex] PV=(m/M)RT [/tex]
    => [tex] P/e=(RT)/M [/tex]

    In both the cases, the length of the pipe is same. Find the relation between the wavelength and the length of the pipe. You should now be able to calculate the fundamental frequency in the second case
  4. Jun 7, 2005 #3
    Probably a stupid question, but what is e? We never learnt this equation.
  5. Jun 7, 2005 #4
    In siddharth's notation, [itex]e[/itex] denotes the density of the gas. That is, [itex]e = \frac{m}{V}[/itex].
  6. Jun 7, 2005 #5
    Okay, here is my approach. Let me know if you think it makes sense.

    We have [itex]c = f\lambda[/itex] and [itex]c = \sqrt{(\gamma RT)/M}[/itex]. In the fundamendal mode, [itex]\lambda = 2L[/itex]. So

    [itex]2f_\mathrm{He}L = \sqrt{(\gamma_\mathrm{He}RT)/M_\mathrm{He}}[/itex] (1)


    [itex]2fL = \sqrt{(\gamma_\mathrm{air}RT)/M_\mathrm{air}}[/itex] (2)

    Dividing (1) by (2),

    [itex]f_\mathrm{He}/f = \sqrt{(\gamma_\mathrm{He}M_\mathrm{air})/(\gamma_\mathrm{air}M_\mathrm{He})}[/itex]


    [itex]f_\mathrm{He} = f\sqrt{(\gamma_\mathrm{He}M_\mathrm{air})/(\gamma_\mathrm{air}M_\mathrm{He})}[/itex]
  7. Jun 8, 2005 #6
    Yep, that does make sense when you go through it like that. I've just never approached c = sqrt(γP/e). Thanks for your help.
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