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anonymous

  • 4 years ago

A 512 Hz tuning fork is placed over a long tube that is submerged in a column of water. The speed of sound in the air during this experiment is 343 m/s. The tube is place at one position and resonance is heard. It is raised to a new position and resonance is heard. What is the distance between the two resonances?

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  1. anonymous
    • 4 years ago
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    There are numerous distances. Essentially, any distance equal to the wavelength of any harmonic of the 512 Hz signal will create a resonance.

  2. anonymous
    • 4 years ago
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    D:

  3. anonymous
    • 4 years ago
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    I feel so stupid i figured it out...just realized the answer was in centimeters -_-. So if anyone wants to see my answer it's v=(wavelength)(frequency) and d = (1/2)(wavelength) \[343 m/s \div 512s\] = .66992 m = wavelength .66992m(1/2) = .3349m = distance convert that to cm it's 33.5 cm ugh physics

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