An open vertical tube is filled with water, and a tuning forkvibrates over its m

Question

An open vertical tube is filled with water, and a tuning forkvibrates over its mouth. As the water level is lowered in the tube,resonance is heard when the water level has dropped 17 cm, andagain after 49 cm of distance exists from the water to the top ofthe tube. What is the frequency of the tuning fork? *I don't even know where to start!! Please Help!!!!! An open vertical tube is filled with water, and a tuning forkvibrates over its mouth. As the water level is lowered in the tube,resonance is heard when the water level has dropped 17 cm, andagain after 49 cm of distance exists from the water to the top ofthe tube. What is the frequency of the tuning fork? *I don't even know where to start!! Please Help!!!!!

Explanation / Answer

the frequency of the tuning fork is n = v / [ 2 ( L ' -L ) ] where v = speed of sound = 343 m / s            L= Firat resonant length = 17 cm = 0.17 m            L' = second resonant length = 49 cm = 0.49 m So, n = 535.93 Hz

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