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.
Q48.
This question is best done graphically as it would have been done in WW1 in France. Decide on a scale so that a line 2000 m long will fit on the graph paper. Suggest 10 cm = 1000 m ( a scale of 1:10000).

Assume the three microphones A, B, C are on a straight line 1000 m apart as shown below. The gun must be a bit further away from B than A as the sound is first heard at A, then B, then C. Draw the gun in position as shown with A, B and C 10 cm apart. You cannot draw in the position of the gun as that is what you are trying to find. The sound radiates out from the gun in a spherical wavefront (or circular if drawn on paper in 2-D) as shown.

The sound takes 1.2 seconds to travel to B after being heard at A.  Assume that the speed of sound in air is 330 m s-1, this corresponds to a distance of 330 x 1.2 = 396 m. If we draw a circle centered at B with a radius of 396 m (3.96 cm) then all points on that line represent the possible locations of a  sound that is 396 m away from B at the same time.

The question also says that the sound at C is heard 3.7 s after it is heard at A. In this time a sound will have travelled 330 x 3.7 = 1221 m. So draw a circle centered on C with a radius of 12.21 cm (= 1221 m at a scale of 1:10000). Mark the intersection as point X.

The sound of the gun is heard at point X at the same time as point A so draw a circle centered at X with a radius of XA. Then draw a circle centered at A with the same radius. Where these lines intersect is the location of the gun G. The distance from the gun to A (GA) is measured as 9 cm ( = 900m). The angle BAG is 82°. The angle BCG is about 25°.


The question implied that there may be more than one solution. This is correct. An imaginary source of sound (in a mirror-image position on the opposite side of the line of microphones as in the diagram below) would also give rise to similar conditions at the microphones. This imaginary source would be ruled out by common sense.

    

Algebraic method


On the diagram above, the circular wavefront is shown as it reaches microphone A. Three lines are drawn from the Gun to microphones A, B and C. As well, a perpendicular "y" is dropped from the Gun to the line of the microphones. It is "x" metres from A. We have three right angled triangles each sharing a common side (y). Three simultaneous equations can be proposed with the two unknowns x and y.

Equation 1 for DGCX:  y2 = (1221 + s)2 - (2000 - x)2

Equation 2 for DGBX:  y2 = (396 + s)2 - (1000 - x)2

Equation 3 for DGAX:  y2 = s2 - x2

Equating Eq 1 and 3:
12212 + 2442s + s2 - 20002 + 4000x  +x2  = s2 - x2

x = 42 m

Equating Eq 2 and 3:
3962 + 792s + s2 - 10002 + 2000x - x2 = s2 - x2

s = 958 m

The angle GAX = cos-1 (x/s) = cos-1 (42/958) = 87.5°

 

 

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