A piano tuner stretches a steel piano wire with a tension of 765 N . The steel wire has a length of 0.600 m and a mass of 4.50g .


What is the frequency (f_1) of the string's fundamental mode of vibration?
What is the number n of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to f = 16 kHz?

Answer :

Answer:

266.14532 Hz

60

Explanation:

m = Mass of string = 4.5 g

T = Tension = 765 N

L = Length of string = 0.6 m

f = Frequency = 16 kHz

Linear density is given by

[tex]\mu=\dfrac{m}{L}\\\Rightarrow \mu=\dfrac{4.5\times 10^{-3}}{0.6}\\\Rightarrow \mu=0.0075\ kg/m[/tex]

Fundamental frequency is given by

[tex]f'=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}\\\Rightarrow f'=\dfrac{1}{2\times 0.6}\sqrt{\dfrac{765}{0.0075}}\\\Rightarrow f'=266.14532\ Hz[/tex]

The fundamental frequency is 266.14532 Hz

The number of loops is given by

[tex]n=\dfrac{f}{f'}\\\Rightarrow n=\dfrac{16000}{266.14532}\\\Rightarrow n=60.11753[/tex]

The number of loops is 60.

Cricetus

The Frequency will be "266.14 Hz" and the number of loops are "60".

Given:

Tension,

  • 765 N

Length,

  • 0.600 m

Mass,

  • 4.50 g

As we know,

Linear density, [tex]\mu = \frac{Mass}{Length}[/tex]

By substituting the values, we get

                                [tex]= \frac{4.5\times 10^{-3}}{0.6}[/tex]

                                [tex]= 7.5\times 10^{-3} \ kg/m[/tex]

The fundamental frequency will be:

→ [tex]F = \frac{1}{2L}\sqrt{\frac{T}{\mu} }[/tex]

      [tex]= \frac{1}{2(0.6)}\sqrt{\frac{765}{7.5\times 10^{-3}} }[/tex]

      [tex]= 266.14 \ hz[/tex]

and,

The number of loops will be:

→ [tex]n = \frac{f'}{f}[/tex]

      [tex]= \frac{16000}{266.14}[/tex]

      [tex]= 60[/tex]

Thus the above response is correct.

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