Question

A piano tuner stretches a steel piano wire with a tension of 1070 N . The wire is 0.400 m long and has a mass of 4.00 g . A. What is the frequency of its fundamental mode of vibration?
B. What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to 1.00×10^4 Hz ?

Answer 1

A. 409 Hz

The fundamental frequency of a string is given by:

$f_1=\frac{1}{2L}\sqrt{\frac{T}{m/L}}$

where

L is the length of the wire

T is the tension in the wire

m is the mass of the wire

For the piano wire in this problem,

L = 0.400 m

T = 1070 N

m = 4.00 g = 0.004 kg

So the fundamental frequency is

$f_1=\frac{1}{2(0.400)}\sqrt{\frac{1070}{(0.004)/(0.400)}}=409 Hz$

B. 24

For this part, we need to analyze the different harmonics of the piano wire. The nth-harmonic of a string is given by

$f_n = nf_1$

where $f_1$ is the fundamental frequency.

Here in this case

$f_1 = 409 Hz$

A person is capable to hear frequencies up to

$f = 1.00 \cdot 10^4 Hz$

So the highest harmonics that can be heard by a human can be found as follows:

$f=nf_1\\n= \frac{f}{f_1}=\frac{1.00\cdot 10^4}{409}=24.5 \sim 24$