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Homework Help: Waves in semiclosed tubes

  1. May 13, 2014 #1
    I'm doing a self-guidance type of study, trying to learn my way through a book based on physics and the sound of music. I haven't dabbled with waves since high school.

    I know that these are sequential harmonics. Therefore, given that it's a semiclosed tube and L = length of tube in meters:

    λ1 = 4L
    λ3 = 4L/3
    λ5 = 4L/5
    λ7 = 4L/7​

    I'm having trouble figuring out how to determine exactly which harmonics these are. Converting the frequencies into their respective periods gives me:

    Tn = [itex]\frac{1}{1000 Hz}[/itex] = 0.001000 s
    Tn+2 = [itex]\frac{1}{1400 Hz}[/itex] = 0.0007143 s
    Tn+4 = [itex]\frac{1}{1800 Hz}[/itex] = 0.0005555 s​

    (where n is the nth harmonic)​

    Looking at the basis for harmonics (in semiclosed tubes), I know that if a wave travels a distance of 4L, or an odd-numbered integral quotient of 4L, in a time equal to its own period, a standing wave is formed. In other words, the 1000 Hz wave listed above must travel 4L/n meters in 0.001000 second for a standing wave to form.

    But how can I determine what n is? One cannot simply assume that 1000 Hz is the fundamental frequency. Likewise, as shown by the basic wave formula of v = ƒλ, one cannot calculate wavelength from frequency/period alone.

    I thought about an approach based around velocity but, again, I only know the time component of one oscillation.

    Assuming room temperature (23 °C; 296 K), I know that:

    ƒn = [itex]\frac{86n}{L}[/itex]

    The calculation would be simple, but I don't know how I can calculate n. I apologize for not having more work to show, and I've tried to make up for that by posting my logic, but I'm at a standstill.
  2. jcsd
  3. May 13, 2014 #2


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    Science Advisor
    Homework Helper

    Converting into wavelengths and time periods is not the easiest way to do this.

    If the fundamental frequency of a closed tube is f, what are the frequencies of the harmonics?

    Try to match up that information with the observed frequencies of 1000, 1400, 1800. (Hint: think about the difference between successive frequencies).
  4. May 13, 2014 #3
    Thanks, Aleph. You're right; I was making it too complicated. :approve:

    For closed tubes, ƒn = nƒ1

    So, in response to your question: ƒ2 = 2f, ƒ3 = 3f, ƒ4 = 4f, and so on.

    With the 400 Hz difference between successive frequencies, and knowing that even harmonics are skipped in semiclosed tubes, that must mean the fundamental frequency is 200 Hz. Which means that 1000 Hz is the fifth harmonic, 1400 Hz is the seventh, and 1800 Hz is the ninth.

    Now, given my formula ƒn = 86n/L, I can calculate the length of the tube. - 0.43 m

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