A piano tuner stretches a steel piano wire with a tension of 803N. The wire is 0

Question

A piano tuner stretches a steel piano wire with a tension of 803N. The wire is 0.400 m long and has a mass of 3.00g. What is the frequency of its fundamental mode of vibration and what is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to 1.00 * 10^4 Hz?

Explanation / Answer

The resonance frequencies of a string of length L that is fixed at both ends are given by: f_n = n*v/(2*L) where v is the wave speed, and n is a positive integer > 0. The fundamental frequency is given by the frequency corresponding to n = 1, and values of n > 1 are the harmonics. The wave speed is given by: v = sqrt(T/(M/L)) where T is the tension, M is the total mass of the string, and L, as before, is the length. The quantity M/L is often called the linear mass density. In this case, we have that: v = sqrt((803 N)/((3.*10^-3 kg)/(0.4 m)) v = 319.374 m/s So the resonance frequencies of the string are given by: f_n = n*(319.374 m/s)/(2*0.5m) f_n = n*3.19*10^2 Hz The fundamental frequency (or first harmonic) is when n = 1, so the fundamental mode has a freqquency of 319 Hz (3.19*10^2 Hz) To calculate the highest harmonic that is audible to a person who can only hear frequencies up to 7825 Hz, we divide 7825 Hz by 319 Hz, and take the integer part of the result: (7825 Hz)/(319 Hz) = 24.5, so the person could hear the 24th harmonic (7665 Hz). The 25th harmonic, at 7984 Hz would be above the frequency threshold of their hearing.

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