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Zorluk: KolayResononace, Vibrating Strings, and Air Columns in Pipes

A tube closed at one end has a length of 0.85 m0.85\text{ m}. If the speed of sound in air is 340 m/s340\text{ m/s}, what is the fundamental frequency of the sound wave produced in the tube?

  1. 100 Hz100\text{ Hz}Cevap
  2. B
    200 Hz200\text{ Hz}
  3. C
    400 Hz400\text{ Hz}
  4. D
    289 Hz289\text{ Hz}

Cevap

100 Hz100\text{ Hz}
For a pipe closed at one end, the fundamental frequency is given by f=v4Lf = \frac{v}{4L}. Substituting v=340 m/sv = 340\text{ m/s} and L=0.85 mL = 0.85\text{ m} yields f=3404×0.85=3403.4=100 Hzf = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100\text{ Hz}.

Adım Adım Çözüm

1
Determine the relationship between pipe length and wavelength for the fundamental mode of a closed pipe.
For a pipe closed at one end, λ=4L=4×0.85 m=3.4 m\lambda = 4L = 4 \times 0.85\text{ m} = 3.4\text{ m}.
A closed tube forms a node at the closed end and an antinode at the open end, corresponding to one quarter of a full wavelength.
2
Calculate fundamental frequency using the wave equation v=fλv = f\lambda.
f=vλ=340 m/s3.4 m=100 Hzf = \frac{v}{\lambda} = \frac{340\text{ m/s}}{3.4\text{ m}} = 100\text{ Hz}.
Frequency equals wave velocity divided by fundamental wavelength.

Anahtar Kavram

Fundamental frequency of a pipe closed at one end
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