Figure a shows a simple apparatus for demonstrating resonance in a tube. A long
ID: 1514186 • Letter: F
Question
Figure a shows a simple apparatus for demonstrating resonance in a tube.
A long tube open at both ends is partially submerged in a beaker of water, and a vibrating tuning fork of unknown frequency is placed near the top of the tube. The length of the air column, L, is adjusted by moving the tube vertically. The sound waves generated by the fork are reinforced when the length of the air column corresponds to one of the resonant frequencies of the tube. Suppose the smallest value of L for which a peak occurs in the sound intensity is 8.71 cm. Take the speed of sound to be 344 m/s.
(a) With this measurement, determine the frequency of the tuning fork. Hz
(b) Find the wavelength and the next two air-column lengths giving resonance. ? = m L2 = m L3 = m
An unknown gas is introduced into the aforementioned apparatus using the same tuning fork, and the first resonance occurs when the air column is 5.63 cm long. Find the speed of sound in the gas. vsound = m/s
IMT 1/4 3/4 First (third harmonic) Water Third resonance (fifth harmonic) (a) Apparatus for demonstrating the resonance of sound waves in a tube closed at one end. The length L of the air column is varied by moving the tube vertically while it is partially submerged in water. (b) The first three resonances of the system.Explanation / Answer
here the pipe is closed at one end and open at other end
fundamental frequency fo = v/2L
fo = 344/(4*0.0871) = 987.4 Hz
(b)
wavelength = 4L = 4*0.0871 = 0.3484 m
second resonating length L2 = 3*lambda/4 = 3*0.0871 = 0.2613 m
third resonating length L3 = 5*lambda/4 = 5*0.0871 = 0.4355 m
++++++++++++++
for unknown gas
f = v/4L
987.4 = v/(4*0.0563)
v = 222.4 m/s <<<<---------answer
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