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    The previous part asked:

    A musical note of several frequencies is sounded at the mouth of a 1.0m long vertical tube that has some water in the bottom. Give the depth of water in the tube if the fundamental frequency heard is 125Hz.

    Which I found to be 0.34m. But the next question just leaves me stumped, no matter what I do nothing seems to work.

    When the first overtone is played in D5.15, at what height will the particles' displacement be out of phase and have the same amplitude as that of the particles 8.0cm above the surface of the water? Give your answer as a distance above the surface of the water.

    Can anybody give me an idea of what I'm meant to do
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    The first harmonic is the when the next stationary wave occurs. This is when there are 3/4 of a wavelength in the tube. You are trying to find the distance from the surface of the water where a particle is in antiphase with a particle 8cm from the surface. The two particles will be half a wavelength apart.Half a wavelength is [(0.66/0.75) *0.5], then 8cm 1/2 wavelength = 0.52m. 0.52m is the distance you are looking for.
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    Thank you, it makes so much sense now
 
 
 

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