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 .

       So for a 1 m/s change:  345m/s = 344m/s  + 0.6(T - 20)  =>  1 m/s = 0.6 (T - 20)
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 [change in v] = 0.6 m/s per degree Celsius.  We're asked for the temperature change producing a velocity change of 1.7 m/s.


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 ˝ %.

 

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 wave amplitude then must be 0.003 x 105  N/m2  = 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

15.  w = mg  = 490 N  => m = 490 N/ 9.8 m/s2 =  50 kg

Chapter 4

2.   Assume that the pillar is 1 foot square, or about 30 cm wide.  
      Thus for wavelength 30 cm,  f = (344 m/s)/ 0.3 m = 1147 Hz  (close to ‘treble’ C).
      Sounds of lower frequency (longer wavelength) will diffract around the pillar and be heard without difficulty. The shorter wavelengths (frequencies ~3 or more
      times higher than 1200 Hz) will be ‘blocked’ by the pillar and not heard clearly except perhaps by reflections.  That's why the 'tweeters' [high
      frequency loudspeakers] in stereo cabinets often are not covered by a grille.

6.  f1~V/v = 0.12  => V = 0.12 v  = 0.12 x 344 m/s = 41.28 m/s…this is about 148 km/hr or almost 92 mph!

9.    If f1 = 172 Hz, then λ= 344/172 = 2 m.  Since path length for both direct and reflected sound is a multiple of 2 m, we have constructive interference at L.
For f2 = 86 Hz, λ = 344/86 = 4m.  The direct sound path of 2 m will be out of phase (there's half a wavelength difference) with the 4m reflected path so destructive interference will result.

10.    f1 – f2 = 523 Hz – 520 Hz = 2.4 Hz

11.    The three fundamental frequencies 'mix' to create three beat frequencies.

         If f1 = 440 Hz,  f2 = 438 hz,  and f3 = 443 Hz, we get:

         f3 – f1 = 3 Hz;    f3 – f2 = 5 Hz;      and f1 – f2 = 2 Hz .  [Note that by convention we always subtract the smaller from the larger.]

Chapter 5.

 

1. [200J/ 4s]/ 5 m2  = 50W/ 5 m2 = 10 W/ m2

 

3.     Since I = P/S,  P = IS = [ 10 -6 W /m2 ] [7 x 10-5  m2 ]  =  7 x 10-11 W  or 70 picowatts 

 

            Over an elapsed time of 10 s, E = Pt = 70 pJ/s x 10 s = 700 pJ or 0.7 nanoJoules

 

5.  From Table 5.2, 1000 W/ m2 => 150 dB.

 

 Or calculate it:  I/Io = 103/10-12 = 1015  ; SIL = 10 log 1015  = 10 (15) = 150 dB

 

 

8.   Each added violin increases Io by one more of itself, hence 2 violins = 2Io;  3 violins => 3 Io, etc.


So [using Table 5.1 ], 3 violins => I/Io = 3  => 75 dB + 5 dB = 80 dB ;

5 violins => I/Io = 5  => 75 dB + 7 dB =  82 dB ;

Similarly, 10 violins => 10  => 85 dB

 

SIL = 95 dB => 75 dB + 20 dB, so how many violins are required to produce this 20 dB increase? From Table 5.1, 20 dB => 1/Io = 100, or 100 violins!

Similarly, 105 dB is (105 – 70) = 30 dB difference => I/Io = 1000!!!

If this seems like a ridiculously high number, remember that 105 dB is much louder than a full orchestra (which includes the loud brasses) playing fff !

 

14.   115 dB – 75 dB = 40 dB difference => intensity ratio of 10 000, or 104  (Table 5.1) 

  Since (intensity ratio) = (amplitude ratio)2  , amplitude ratio = square root (10 000) = 100.




Chapter 6


7.  jnd = 2 dB = 10 log I/Io;   2 dB => 1.6, from Table 5.1 .  Thus the difference I - Io is (1 – 1.6) = 0.6  = 60%

 

8.  (300 Hz)/2 = 150 Hz ; (500 Hz) x 4 = 2000 Hz

 

9.  (a) 400 Hz/ 180 Hz = 2.22    ; (b)  7000 Hz/ 400 Hz = 1.75

Frequency ratio (a) is more than an octave; (b) is less than an octave, so (a) will be perceived as the greater change in pitch.


11.  Use the curve in Figure 6.13 and a straightedge or ruler to find the loudness-level  corresponding to the given loudness:

(a) 0.5 sone   ŕ  ~35 phon

(b)   4 sone   ŕ ~ 60 phon

(c)  25 sone   ŕ  ~80 phon

 

17.  From Figure 6.13, 40 phon is ~ 0.8 sone, and is equivalent to 40 dB SIL at 1 kHz.

Boosting the SIL by 40 dB over the whole range of frequencies is a factor of 10,000 and puts the SIL at 80 dB
[which js "factory noise" level, see Table 5.2 !]. So the neighbors will no doubt complain about the racket! 
Note that the extreme bass and treble frequencies will seem even louder, some by 10 phons!


 
 

Chapter 10

 

1.    L = 0.6 m;  l = 2L/n . l1 = 2(0.6 m)/ 1 = 1.2 m ; l2 = 2(0.6 m)/ 2 = 0.6 m        l3 = 2(0.6 m)/ 3  = 0.4 m

 

5. fn = n/2L [T/m]1/2    so doubling T increases f by [2] 1/2  = 1.414

Each semitone is an increase of [2] 1/12  so [2] 1/2  =  { [2] 1/2}}6 or 6 semitones.

 

7.  fn = n/2L [T/m]1/2 = 1/ 2 (0.75m) [ 540/0.006]1/2 = 200 Hz

 

8. fn = n/2L [T/m]1/2  => T/m = 4 L2f2   => T = 4 L2f2 m = 4 (2m)2 (30 Hz)2 (0.06kg.m) = 864 N

 

12. The shorter string is stiffer since it has less length over which it can flex.  Thus it's more like a rod and hence has greater inharmonicity.

 

20. The absence of harmonics near n = 9, 10 indicates that there is at least one node in this vicinity.  Hence it must have been plucked at about half of L/9 from the end, or about L/18. Odd multiples of L/18 would work also: 3/18L = L/6, etc.

 

21. For mode 3, we want nodes at L/3 from either end; thus the antinodes will be at the center and at L/6 from the ends.  So pluck the string at L/6, or at the middle.  To kill f1 and f2, touch the string at the f3 node, L/3 from end.

This harmonic is the "twelfth",  or (octave + a fifth).  To visualize this, look at the harmonic chart (inside text back cover).  If f1were C3 (130.8 Hz),
f3 is 3 x 130.8 Hz = 392.4 Hz = G4, etc.

 

************************************************************************

 

Chapter 11

 

5.  G3/E1 = 196/41,2 = about 4.75 .  So, a double bass  'scaled up' from a violin would have:

body width  21 cm x 4.75 = 1 m; and total length would be ~60 cm x 4.75 = 2.8 m (That's about 9 feet!!) .   You'd need a huge 'reach' to be able to
finger the notes on a 'bull fiddle' this tall.

 

7.  f1 = 1/2L 2L [T/m]1/2  => m = T/4 L2f2 = 60/ 4(0.35m)2 (440)2 = 0.63 grams/ m

 

15. The goblet is more likely to be shatterable when empty.  Liquid in it effectively makes the walls stiffer. More importantly, wall motion moves the liquid which absorbs much of the energy that otherwise goes into making the walls vibrate. [To shatter it, the vibration amplitude would have to exceed the elastic limit of the glass].

 

20.   The ratio between semitones is 21/12 = 1.05946.  For a change of one semitone on a string instrument,  since f is proportional to 1/L, L is shortened to L/1.05946 = 0.94387L [or lengthened to 1.05946 L]. 

The change in length  is  (1 - 0.94387) = 0.05613L .

 

On the violin, at the lower end of the string you'd move your finger 33 cm x  (0.05613) = 1.85 cm per semitone, or about a finger width.

 

On the double bass, however, a semitone change in pitch would require the string to be shortened/lengthened by  104 cm x  (0.05613) = 5.8 cm

Moving up a fourth (5 semitones) in pitch would be a move of 5 x 5.8 cm = nearly 30 cm before you can go to the next higher string.  If the strings were tuned in fifths you'd have to move 7 x 5.8  = over 40 cm. The instrument would be difficult to play unless you had exceptionally long reach.

 

 
 

23.   f1 = 1/(2L)  [T/µ]1/2 . Since µ = m/L,  using this and solving for T gives  T = 4 Lf2 m .  Keeping L and  m constant, a change in f requires that T2/T1 =  (f2/f1)2   .

Thus T2 = T1 (440/415)2 = 1.12T1 . So T was increased by (1.12 - 1) or about 12 % .

 

Changing the length while maintaining the higher pitch: T2/T1 = [L2ff2/L1f1]2 = 1.195,

which is a change of (1.195 -1) = 20%

 

If you changed the mass of the string,  T2/T1 = f22 m2 / f12 m1 =  f22 (1.1m1 / f12 m1

=   212960/172225 = 1.236. or about 24% .

 

If all of the changes were made at once, we have to include all of them in the calculation.

The result: the changes are multiplicative, not additive!

So 1.12 x 1.2 x 1.24 = 1.67 which is a net change in tension of 67%  !!


 

************************************************************************


 

Chapter 12

 

1. Displacement antinodes correspond to locations where the molecular motion of the air is greatest, i.e., where the air is least confined.  Conversely, pressure antinodes (maximum pressure!) occur where the air is most confined.

Thus maximum pressure <--> minimum displacement

 

5. For the open tube, ln = 2L/n .   So l1  = 17 cm x 2 = 34 cm  =>

 f1 = (344m/s)/0.34m  = 1011 Hz

Similarly,  f5 = 344/(34/5) = 5058 Hz

 

7.  Open : fn = nv/2L  => L = nv/2f = (344 m/s)/ 2(860) = 0.22m or about 20 cm

 

            closed: fn = (2n - 1)/4L  =>  L = 344/4(860) = 0.1 m or about 10 cm

 

12.  Higher harmonic modes in open pipes weaken rapidly above N = L/D [see text, p. 246, near the top of the page].

So for L= 40 cm and D = 10 cm  very few harmonics are likely above N= 40/10 = 4

However, for D = 2 cm , N =  L/D = 40/2 = 20....so many more harmonics are likely.

 

15. We will now have a displacement antinode at the hole, so  the pipe acts like an open pipe with L =  44 cm.  Thus f = v/2L = (344m/s)/ 0.88m = 391 Hz [close to G4].

 

20.  Standing wave nodes are  l/2 apart......review Figures 10.1  and 12.2 if you aren't sure about this!  So an adjoining antinode is halfway between, or l/4 away.

 

 

22.  The 'fat' pipe will have fewer harmonics [see #12, above], hence a 'duller', 'flutier' timbre.  The thin pipe has more harmonics and thus might have some 'stringy' quality.

However, the timbre is affected by several other parameters: height and position of the pipe mouth;s upper lip, lip thickness, windway opening, wind pressure, etc.

 

************************************