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An open vertical tube has water in it. a tuning fork vibrates over its mouth. as the water level is lowered in the tube, a resonance is heard when the water level is 180 cm below the top of the tube, and again after the water level is 220 cm below the top of the tube a resonance is heard. what is the frequency of the tuning fork? the speed of sound in air is 343 m/s. answer in units

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Answer:

[tex]428.75\ \text{Hz}[/tex]

Explanation:

[tex]\Delta y[/tex] = Change in water level = [tex]220-180=40\ \text{cm}[/tex]

[tex]\lambda[/tex] = Wavelength

[tex]v[/tex] = Speed of sound = 343 m/s

Between the points of resonance there exists [tex]\dfrac{1}{2}\lambda[/tex]

[tex]\dfrac{1}{2}\lambda=\Delta y\\\Rightarrow \lambda=2\Delta y\\\Rightarrow \lambda=2\times 40\\\Rightarrow \lambda=80\ \text{cm}[/tex]

Wavelength is given by

[tex]f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{343}{0.8}\\\Rightarrow f=428.75\ \text{Hz}[/tex]

The frequency of the tuning fork is [tex]428.75\ \text{Hz}[/tex].

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