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

Question

A piano tuner stretches a steel piano wire with a tension of 803 . The wire is 0.400 long and has a mass of 3.00 . Part A What is the frequency of its fundamental mode of vibration? (Hz) Part B 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

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)/((3.00*10^-3 kg)/(0.4 m)) v = 327.210 m/s So the resonance frequencies of the string are given by: f_n = n*(327.210 m/s)/(2*0.4m) f_n = n*409.01 Hz - Answer for Part - A The fundamental frequency (or first harmonic) is when n = 1, so the fundamental mode has a freqquency of 409.01 Hz To calculate the highest harmonic that is audible to a person who can only hear frequencies up to 1*10^4 Hz, we divide 1*10^4 Hz by 409.01 Hz, and take the integer part of the result: (1*10^4 H Hz)/(409.01 Hz) = 24.44 rounded to 25 - Answer for B

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