Constants Part A A piano tuner stretches a steel piano wire with a tension of 76

Question

Constants Part A A piano tuner stretches a steel piano wire with a tension of 765 N. The steel wire has a length of 0.700 m and a mass of 5.25 g What is the frequency fi of the string's fundamental mode of vibration? Express your answer numerically in hertz using three significant figures View Available Hint(s) Hz Submit Part B What is the number T 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. View Available Hint(s)

Explanation / Answer

Part A.

Speed of a wave on string is given by:

V = sqrt (T/u)

u = linear density = 0.00525 kg/0.700 m = 7.5*10^-3 kg/m

T = 765 N

V = sqrt (765/(7.5*10^-3)) = 319.37 m/sec

Now fundamental frequency is given by

f1 = V/2L

f1 = 319.37/(2*0.700) = 228.12 Hz

Part B

We need to find number of antinodes, So

nth harmonic frequency = frequency/fundamental frequency

n = 16000/228.12 = 70.14

So it can complete the 70th Harmonic

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