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Response Details:
A 107-cm-long pipe is stopped at one end. Near
the open end, there is a loudspeaker that is driven by an audio
oscillator whose frequency can be varied from 10.0 to 4600 Hz. (Take the speed of sound to be 343 m/s.)
(a) What is the lowest frequency of the oscillator that will produce resonance within the tube?
lowest frequency=v/4l=343/4*1.07=80.14=80Hz(approximate)
(b) What is the highest frequency that will produce resonance?
next higher frequencies will be in the ration 1:3:5:7etc
hence
the highest frequency =57*80=4560Hz
(c) How many different frequencies of the oscillator will produce resonance? (Neglect the end correction.)
1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53,55,57
so 29
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