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A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length L is 42.5 cm , 58.6 cm , and 74.7 cm
What is the frequency of the tuning fork? Assume vsound = 343 m/s

Respuesta :

Answer:

The frequency of the tuning is 1.065 kHz

Explanation:

Given that,

Length of tube = 40 cm

We need to calculate the difference between each of the lengths

Using formula for length

[tex]\Delta L=L_{2}-L_{1}[/tex]

[tex]\Delta L=74.7-58.6[/tex]

[tex]\Delta L=16.1\ m[/tex]

For an open-open tube,

We need to calculate the fundamental wavelength

Using formula of wavelength

[tex]\lambda=2\Delta L[/tex]

Put the value into the formula

[tex]\lambda=2\times16.1[/tex]

[tex]\lambda=32.2\ cm[/tex]

We need to calculate the frequency of the tuning

Using formula of frequency

[tex]f=\dfrac{v}{\lambda}[/tex]

Put the value into the formula

[tex]f=\dfrac{343}{32.2\times10^{-2}}[/tex]

[tex]f=1065.2\ Hz[/tex]

[tex]f=1.065\ kHz[/tex]

Hence, The frequency of the tuning is 1.065 kHz

Cricetus

The frequency of the tuning will be "1.065 kHz".

Tuning fork:

According to the question,

Length of tube = 40 cm

Difference between each of the lengths will be;

ΔL = [tex]L_2-L_1[/tex]

    = [tex]74.5-58.6[/tex]

    = [tex]16.1 \ m[/tex]

We know,

→ λ = 2ΔL

By substituting the values,

       = [tex]2\times 16.1[/tex]

       = [tex]32.2 \ cm[/tex]

hence,

The frequency will be:

→ [tex]f = \frac{v}{f}[/tex]

      [tex]= \frac{343}{32.2\times 10^{-2}}[/tex]

      [tex]= 1065.2 \ Hz[/tex]

      [tex]= 1.065 \ kHz[/tex]

Thus the above answer is correct.

Find out more information about the frequency here:

https://brainly.com/question/1512031

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