a piano note is compared to a tuning fork vibrating at 440 Hz. Four beats per se

Question

a piano note is compared to a tuning fork vibrating at 440 Hz. Four beats per second are discerned by the piano tuner. When the tension in the string is increased slightly, the beat frequency increases. What was the initial frequency of the piano wire?

Explanation / Answer

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((765 N)/((3.75*10^-3 kg)/(0.5 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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