Senior Physics - Solutions
Chapter 16. Sound, Audio Technology - Worked Solutions
...to selected Complex Reasoning questions from the OUP text "New Century" Senior Physics - Knowledge, Processes and Reasoning by Walding, Rapkins and Rossiter
Resonance is heard when frequency produced by the air column is the same as the tuning fork.
From case 1 to 2, or between 2 and 3 there is ˝ a wavelength.
therefore as in one case the length is 12 cm and in the other it is 50cm,
then (50 - 12) cm is equal to ˝ l.
That is l = 76 cm
since v = f l
then f = v / l = 334 /0.76m = 439 Hz
fx = 245 Hz, and fy = 247 Hz
Z with x gives 3 beats per second. Therefore fz is either 245+ 3 = 248 Hz
or 245 - 3 = 242 Hz
Z with y gives 1 beat per second. Therefore fz is either 247 + 1 = 248 Hz or 247 - 1 = 246 Hz.
The common answer is fz = 248 Hz.
Tuning fork’s frequency = 442 Hz
With the ‘A ‘ string a beat frequency of 5 Hz is heard.
Therefore the ‘A’ string is either 442 + 5 = 447 or 442 - 5 = 437 Hz.
With the rubber band the beat frequency of 3 Hz is heard. Since by adding the rubber band slows the vibrations and thus the frequency. The fork will be less. If the ‘A’ string is 447 then the beat frequency will be greater as frequency of the tuning fork will be further from the string when a rubber band is added.
Since it is closer to the string’s frequency ( the beat frequency is now 3 Hz) the frequency of the ‘A’ string is 437 Hz.
Since the source of the sound is moving towards the cliff the frequency of the sound measured at the cliff will be
10 kmh-1 = 2.8 ms-1.
f1 = f v/(v - vs) = (480 x 330)/(330 - 2.8) = 484.1 Hz
However the cliff reflects this sound back to a moving observer the frequency of the sound measured by this observer will be:
f1 = f(v + v0)/v
Therefore f1 = 484.1 (330 + 2.8)/330 = 488.2 Hz
f = 440 Hz, with the train moving towards the station.
The person at the station would hear a frequency of f1 = f(v/(v - vs)
and will be of high frequency.
Since there is a beat frequency of 3 the observer on the station would measure the frequency of the trumpet on the moving train to be (440 + 3) Hz = 443 Hz.
solving the equation :-
443 = 440x(340 /(340 - vs))
calculates to vs = 2.3 ms-1. or 8 kmh-1.
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