An open vertical tube has water in it. A tuning fork vibrates over its mouth. As

Question

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 19.25 cm
below the top of the tube. And again, after
the water level is 22.75 cm below the top of
the tube a resonance is heard.
The speed of sound in air is 343 m/s.
What is the frequency of the tuning fork?
Answer in units of Hz. How many nodes are in the tube afterthe
water has reached the second distance from
the top of the tube?

Explanation / Answer

The key idea is that the difference in distance is half awavelength. So... . (1/2) = 22.75 -19.25          = 7.00 cm = 0.0700 m. . Then... .       f = v / = 343 / 0.0700 =    4900Hz . The number of half wavelengths in the tube is... .              22.75 / 3.50 =   6.5      so there are   seven nodes in thetube including the one at the top of the tube

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