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3 years ago

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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 th

e 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

1 answer:

Umnica [9.8K]3 years ago

3 0

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

$\Delta L=L_{2}-L_{1}$

$\Delta L=74.7-58.6$

$\Delta L=16.1\ m$

For an open-open tube,

We need to calculate the fundamental wavelength

Using formula of wavelength

$\lambda=2\Delta L$

Put the value into the formula

$\lambda=2\times16.1$

$\lambda=32.2\ cm$

We need to calculate the frequency of the tuning

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{343}{32.2\times10^{-2}}$

$f=1065.2\ Hz$

$f=1.065\ kHz$

Hence, The frequency of the tuning is 1.065 kHz

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