in as chapter- 2708744 Signed es session masteringphysics.com/myct/itemview Reso
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in as chapter- 2708744 Signed es session masteringphysics.com/myct/itemview Resources previous l 13 of 14 l ne PeysosiFa 2014 Vemuru t Harmonics a Piano Wine of What is the frequency fi the string's fundamental mode of vibration? of Part A Express your answer numerically in hertz using three significant figures. t Harmonics of a Piano Wire A piano tuner stretches a steel piano wire with a tension of 765 N. The steel wire has length of a O and a mass of 6.75g fi 117 Hints My Answers Give Up Review Part Incorrect; Try Again: 17 attempts remaining Part B What the number n ofthe highest harmonic that oould be heard by a person who is capable of hearing frequencies up to f 16 kHz? Express your answer exactly.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((765 N)/((6.75*10^-3 kg)/(0.9 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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