piano tuner

2012-11-24 11:03 pm

A piano tuner stretches a steel piano wire with a tension of 765 N.
The steel wire has a length of 0.900m and a mass of 6.75g .

1. What is the frequency of the string's fundamental mode of vibration?

Express your answer numerically in hertz using three significant figures.

2. What is the number n of the highest harmonic that could be heard by
a person who is capable of hearing frequencies up to f = 16 kHz?
Express your answer exactly.

回答 (1)

天同

2012-11-25 12:35 am

✔ 最佳答案

1. Mass per unit length of the piano wire = 6.75 x 10^-3/0.9 kg/m = 7.5 x 10^-3 kg/m
Using equation: v^2 = T/p
where v is the speed of wave on the string, T is tension of the string and p is the mass per unit length of the string.
hence, v^2 = 765/7.5x10^-3 (m/s)^2
v = 319.37 m/s

At the fundmaental mode of vibration, wavelength = 2 x 0.9 m = 1.8 m
Hence, frequency of vibration = 319.37/1.8 Hz = 177 Hz

2. At 16 kHz, wavelength = 319.37/16000 m = 0.01996 m
No. of wavelengths formed on the wire = 0.9/0.01996 = 45
Hence, the no. of harmonics = 45 x 2 = 90

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