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

A sonometer wire of length 0.80 m0.80\text{ m} and mass 2.0 g2.0\text{ g} is maintained under a tension of 100 N100\text{ N}. What is the fundamental frequency of vibration of the wire?

  1. 125 Hz125\text{ Hz}Cevap
  2. B
    250 Hz250\text{ Hz}
  3. C
    62.5 Hz62.5\text{ Hz}
  4. D
    12.5 Hz12.5\text{ Hz}

Cevap

The fundamental frequency of vibration of the wire is 125 Hz125\text{ Hz}.
The linear mass density is μ=0.002 kg0.80 m=0.0025 kg/m\mu = \frac{0.002\text{ kg}}{0.80\text{ m}} = 0.0025\text{ kg/m}. The velocity of waves on the string is v=1000.0025=200 m/sv = \sqrt{\frac{100}{0.0025}} = 200\text{ m/s}. The fundamental frequency is f=v2L=2002×0.80=125 Hzf = \frac{v}{2L} = \frac{200}{2 \times 0.80} = 125\text{ Hz}.

Adım Adım Çözüm

1
Convert mass to kilograms and calculate linear mass density (μ)(\mu).
m=2.0 g=0.002 kgm = 2.0\text{ g} = 0.002\text{ kg}. Thus, μ=mL=0.002 kg0.80 m=0.0025 kg/m=2.5×103 kg/m\mu = \frac{m}{L} = \frac{0.002\text{ kg}}{0.80\text{ m}} = 0.0025\text{ kg/m} = 2.5 \times 10^{-3}\text{ kg/m}.
Standard SI units must be used for tension in Newtons and length in meters.
2
Calculate the speed of the transverse wave on the string (v)(v).
v=Tμ=1000.0025=40000=200 m/sv = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.0025}} = \sqrt{40000} = 200\text{ m/s}.
Wave speed on a stretched string depends directly on tension and inversely on linear density.
3
Calculate the fundamental frequency (f1)(f_1).
f1=v2L=2002×0.80=2001.6=125 Hzf_1 = \frac{v}{2L} = \frac{200}{2 \times 0.80} = \frac{200}{1.6} = 125\text{ Hz}.
For a fixed string vibrating in its fundamental mode, the length equals half the wavelength (L=λ2\,L = \frac{\lambda}{2}\,).

Anahtar Kavram

Fundamental frequency of a stretched string
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