Chapter 16 - Worked Solutions
to selected extension questions from the OUP text Senior Physics -
Concepts in Context by Walding,
Rapkins and Rossiter
Q43.
Resonance is heard when frequency produced by the air column is the same as the tuning fork.
The diagrams show the two simplest cases where standing waves will form. The
difference between Diagram 1 and 2 is 1/2 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.
PLEASE NOTE: students will often say Diagram 1 represents 1/4 l
so l = 48 cm. We
say that there is no indication of end correction to explain the anomaly.
However, in subsequent editions we will modify this question because if l
is 76cm then 1/4l
is 19 cm which is more than the question says. We'll change it to 21cm and 59cm
thus allowing for a 2cm end correction. We're not sure why we're telling you
this but it is cathartic.
Q44.
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.
Q 45.
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.
Q 46.
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
Q47.
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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