PHY2464 Homework Solutions
Chapter 1
1. v = d/t => d = tv = 60 s x 344 m/s = 20,640 m or 20.6 km
For t = 1 hr, d = 20.6 km/min (from above) x 60 m.hr = 1240 km
For t = 0.002 s,
d = 344 m/s x 2 x 10 ` 3 s = 688 x 10 `3 m = 68.8 cm
4. Given: v = 344 + 0.6 (T – 20) m/s .
This tells us that for every degree Celsius rise in air temperature above 20
degrees, the speed of sound increases by 0.6 m/s.
Thus Δv = 0.6 m/s per degree Celsius.
Therefore (1.7 m/s)/ [0.6 m/s per degree] = 2.8333 degrees, or about 3 degrees
Celsius. Looking at the formula [or “functional relationship” as Hall calls it], this
says that a temperature change of about 3 degrees Celsius either way (up or down)
will result in a velocity change, and hence a frequency or pitch change of around
½ %.
6. For an estimated chest area S = 1’ x 1’ , this is about 30 cm x 30 cm = 0.09 m2 .
Since PATM = 105 N/m2 = F/S => F = 105 N/m2 x 0.09 m2 = 9000N ;
9000N/500N/person = 18 persons (!)
8. 60 lb/in2 / 14.7 lb/in2 per atmosphere = 4.08 atmospheres or about 4 ATM.
4 ATM x 105 N/m2 per ATM = 4 x 105 N/m2 per tire.
To support a total weight of 800N, we have 400N per tire.
Since p = F/S => S = F/p = 400N/ 4 x 105 N/m2
= 0.001 m2 or about 10 cm2
12. Assuming sinusoidal waves, 0.997 ATM for the rarefactions implies a
pressure of 1.003 ATM for the compressions [1.000 – 0.997 = 0.003]. The
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wave amplitude then must be 0.003 x 10 N/m = 300 N/m
This is a very loud sound! See the last paragraph above the Summary on p. 13 .
Chapter 2
2. P = 1.f => f = 1/P = 1/ 5 x 10-3 s = 200 Hz
5. v = fλ = 175 Hz x 2 m = 350 m/s
This differs from 344 m/s; perhaps due to a higher temperature than 20 Celsius, or
maybe due to different composition of the air.
6. λ = v/f = 344 m/s / (1000s) = 0.344 m or 34.4 cm
9. λ = 4 m => f = v/λ = 344 m/s/ 4m = 86 Hz
13. Amplitude = (3.8 m - 3.2 m)/ 2 = 0.3 m or 30 cm
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15. w = mg = 490 N => m = 490 N/ 9.8 m/s = 50 kg
Chapter 4.
2.