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Zorluk: Çok zorResononace, Vibrating Strings, and Air Columns in Pipes

A uniform wire of length 0.60 m0.60\text{ m} and linear mass density 4.0×103 kg/m4.0 \times 10^{-3}\text{ kg/m} is fixed at both ends under a tension of 360 N360\text{ N}. When plucked, the wire vibrates in its second overtone. This frequency is found to be in resonance with the first overtone of an air column in a pipe closed at one end. Taking the speed of sound in air as 340 m/s340\text{ m/s}, what is the length of the pipe?

  1. A
    0.17 m0.17\text{ m}
  2. 0.34 m0.34\text{ m}Cevap
  3. C
    0.51 m0.51\text{ m}
  4. D
    0.68 m0.68\text{ m}

Cevap

The length of the pipe is 0.34 m0.34\text{ m}.
The correct answer of 0.34 m0.34\text{ m} is obtained by finding the wave speed on the string (300 m/s300\text{ m/s}), computing its 3rd harmonic frequency (750 Hz750\text{ Hz}), and equating this to the 3rd harmonic frequency formula for a closed pipe (f=3va4Lpf = \frac{3 v_a}{4 L_p}).

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1
Calculate the speed of transverse waves on the stretched string.
vs=Tμ=360 N4.0×103 kg/m=90000=300 m/sv_s = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{360\text{ N}}{4.0 \times 10^{-3}\text{ kg/m}}} = \sqrt{90000} = 300\text{ m/s}
Wave speed on a stretched string depends on tension and linear mass density.
2
Determine the fundamental frequency and the 2nd overtone frequency of the string.
Fundamental frequency f1,s=vs2Ls=3002(0.60)=250 Hzf_{1,s} = \frac{v_s}{2 L_s} = \frac{300}{2(0.60)} = 250\text{ Hz}. Second overtone is the 3rd harmonic (n=3n=3), so f=3×250=750 Hzf = 3 \times 250 = 750\text{ Hz}.
For a string fixed at both ends, harmonics are integer multiples of the fundamental, and the 2nd overtone corresponds to n=3n=3.
3
Relate the resonance frequency to the length of the closed pipe.
First overtone of a closed pipe is its 3rd harmonic (m=3m=3): f=3va4Lp    750=3(340)4Lpf = \frac{3 v_a}{4 L_p} \implies 750 = \frac{3(340)}{4 L_p}.
Pipes closed at one end only produce odd harmonics (1,3,5,1, 3, 5, \dots), so the 1st overtone is m=3m=3.
4
Solve for the pipe length LpL_p.
Lp=3(340)4(750)=10203000=0.34 mL_p = \frac{3(340)}{4(750)} = \frac{1020}{3000} = 0.34\text{ m}
Algebraic rearrangement yields the required physical length.

Anahtar Kavram

Harmonics and overtones in vibrating strings and closed air columns under resonance
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