A vibrating tuning fork of frequency 512 Hz is held over a water column with one end closed and the other open. As the water level is allowed to fall, a loud sound (resonance) is heard at specific water levels. Assume you start with the tube full of water, and begin steadily lowering the water level. What is the water level (as measured from the top of the tube) for the third such resonance? Take the speed of sound in air to be 343 m/s.
in a fundamental frequency (f):
λ=2L/n and f=n/2L .v for waves whose both ends are open
λ=4L/n and f=n/4L .v for waves whose one end closed and the second is open
n is a positive integer; n= 0,1,2,3,4,5,6....
v is the velocity
L is the length of string
λ is the wavelength
The Attempt at a Solution
here we have one open and other closed ends then we use λ=4L/n and f=n/4L .v
so f=512Hz, n=3 (third such resonance) and v=343m/sec
substitute in the equation
we get L equal 0.502m=50.2cm then it is D
but why the correct answer is A??
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