Questions
A resonance occurs with a tuning fork and an air column of size 12 cm. The next higher resonance occurs with an air column of 38 cm. What is the frequency of the tuning fork? Assume that the speed of sound is 312 m/s.
detailed solution
Correct option is C
Since the standing wave mode has a displacement antinode at the opening, there is a displacement node at the water-air interface. By increasing the height of the air column, to go from one harmonic to the next, an additional length equal ½ of wavelength is required. Henceλ2=(0.38−0.12)m⇒λ=0.52mFinally. from v=fλ, we find that f=v/λ=312/0.52=600HzTalk to our academic expert!
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A tube of certain diameter and of length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz. The velocity of sound in air is 320 m/s. One end of the tube is now closed, considering the effect of end correction, calculate the lowest frequency of resonance for the tube (in Hz).
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