A piano tuner stretches a steel piano wire with a tension of 1020 N . The wire i

ID: 1497003 • Letter: A

Question

A piano tuner stretches a steel piano wire with a tension of 1020 N . The wire is 0.400 m long and has a mass of 3.80 g .

Part A

What is the frequency of its fundamental mode of vibration?

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 1.00×104 Hz ?

A piano tuner stretches a steel piano wire with a tension of 1020 N . The wire is 0.400 m long and has a mass of 3.80 g .

Part A

What is the frequency of its fundamental mode of vibration?

f1 = 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 1.00×104 Hz ?

Here ,

tension of wire , T = 1020 N

L = 0.40 m

mass , m = 3.80 gm

m = 0.0038 Kg

part A)

frequency of its fundamental mode = sqrt(T * L/m)/(2 * L)

frequency of its fundamental mode = sqrt(1020 * 0.40/.0038)/(2 * 0.40)

frequency of its fundamental mode = 409.6 Hz

the frequency of its fundamental mode is 409.6 Hz

part B)

let the number of highest harmonic is N

N * frequency of its fundamental mode = 1 *10^4

N * 409.6 = 1 *10^4

N = 24.4

the number of highest harmonic is 24

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