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show that the frequency of the nthharmonic of a tube of length L that is open at one end and closed at the other end is givenby𝑓𝑛 = 𝑛 ( 𝑣4𝐿 )

Question

Show that the frequency of the nth harmonic of a tube of length L that is open at one end and closed at the other end is given by

f_n = n \left( \frac{v}{4L} \right)

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Solution

Sure, here's how you can show that the frequency of the nth harmonic of a tube of length L that is open at one end and closed at the other end is given by 𝑓𝑛 = 𝑛 ( 𝑣/4𝐿 ):

  1. The fundamental frequency (or first harmonic) of a tube that is open at one end and closed at the other is determined by the length of the tube and the speed of sound in the medium. The wavelength of the fundamental frequency is four times the length of the tube (λ = 4L), because the tube length represents a quarter of a wave.

  2. The frequency of a wave is given by the speed of the wave divided by its wavelength (f = v/λ). Substituting λ = 4L into this equation gives the fundamental frequency: f1 = v/(4L).

  3. The nth harmonic of a wave is simply n times the fundamental frequency. Therefore, the frequency of the nth harmonic is given by 𝑓𝑛 = 𝑛 ( 𝑣/4𝐿 ).

This shows that the frequency of the nth harmonic of a tube of length L that is open at one end and closed at the other end is given by 𝑓𝑛 = 𝑛 ( 𝑣/4𝐿 ).

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