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

An organ pipe open at both ends has a length of 0.60 m0.60\text{ m}. The fundamental frequency of resonance for this pipe is observed to be equal to the frequency of the first overtone of a stretched wire of length 0.40 m0.40\text{ m} fixed at both ends. If the linear mass density of the wire is 4.0×104 kg/m4.0 \times 10^{-4}\text{ kg/m} and the speed of sound in air is 330 m/s330\text{ m/s}, what is the tension in the wire?

  1. A
    1.21 N1.21\text{ N}
  2. 4.84 N4.84\text{ N}Cevap
  3. C
    9.68 N9.68\text{ N}
  4. D
    19.36 N19.36\text{ N}

Cevap

The tension in the wire is 4.84 N4.84\text{ N}.
The fundamental frequency of the open pipe is f=v2L=3302(0.60)=275 Hzf = \frac{v}{2L} = \frac{330}{2(0.60)} = 275\text{ Hz}. Since the wire vibrates in its first overtone (second harmonic, n=2n=2), its frequency is f=2vwire2Lwire=vwire0.40=275 Hzf = \frac{2 v_{\text{wire}}}{2 L_{\text{wire}}} = \frac{v_{\text{wire}}}{0.40} = 275\text{ Hz}. This gives vwire=110 m/sv_{\text{wire}} = 110\text{ m/s}. Using vwire=T/μv_{\text{wire}} = \sqrt{T/\mu}, the tension is T=μvwire2=(4.0×104)(110)2=4.84 NT = \mu v_{\text{wire}}^2 = (4.0 \times 10^{-4})(110)^2 = 4.84\text{ N}.

Adım Adım Çözüm

1
Calculate the fundamental frequency of the pipe open at both ends.
fpipe=vair2Lpipe=3302×0.60=275 Hzf_{\text{pipe}} = \frac{v_{\text{air}}}{2 L_{\text{pipe}}} = \frac{330}{2 \times 0.60} = 275\text{ Hz}.
An open pipe supports a fundamental wavelength of λ=2Lpipe\lambda = 2L_{\text{pipe}}.
2
Express the frequency of the first overtone of the stretched wire and equate it to the pipe frequency.
fwire, 2=2vwire2Lwire=vwireLwire=275 Hzf_{\text{wire, 2}} = \frac{2 v_{\text{wire}}}{2 L_{\text{wire}}} = \frac{v_{\text{wire}}}{L_{\text{wire}}} = 275\text{ Hz}.
For a wire fixed at both ends, the first overtone corresponds to the second harmonic (n=2n = 2).
3
Solve for the speed of the transverse wave on the wire.
vwire=275×0.40=110 m/sv_{\text{wire}} = 275 \times 0.40 = 110\text{ m/s}.
Rearranging f=vwire/Lwiref = v_{\text{wire}} / L_{\text{wire}} gives vwire=fLwirev_{\text{wire}} = f \cdot L_{\text{wire}}.
4
Calculate the tension in the wire using the wave speed formula.
T=μvwire2=(4.0×104)×1102=4.84 NT = \mu v_{\text{wire}}^2 = (4.0 \times 10^{-4}) \times 110^2 = 4.84\text{ N}.
The speed of a transverse wave on a stretched string is given by v=T/μv = \sqrt{T / \mu}.

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