Questions:
1- (a) 2.0; (b) (1.5); (c) fully constructive in (a) ; (d) fully destructive in (b)
6- all odd harmonics
8- E, A, D, C, B
6. Let be the length of the rod. Then the time of travel for sound in air (speed vs) will
be ts  / vs . And the time of travel for compressional waves in the rod (speed vr) will be
tr  / vr . In these terms, the problem tells us that
1 1
ts  tr  0.12s     .
 v s vr 
Thus, with vs = 343 m/s and vr = 15vs = 5145 m/s, we find = 44 m .
16. At the location of the detector, the phase difference between the wave which traveled
straight down the tube and the other one which took the semi-circular detour is
  k d 
2
( r  2r ).

For r = rmin we have  = , which is the smallest phase difference for a destructive
interference to occur. Thus
rmin 

40.0cm

 17.5cm.
2(   2) 2(   2)
30. (a) The intensity is
I
P
30.0 W

 5.97  105 W/m 2 .
2
2
4r
(4)(200 m)
(b) Let A (= 0.750 cm2) be the cross-sectional area of the microphone. Then the power
intercepted by the microphone is
P  IA  0  (6.0  105 W/m2 )(0.750cm2 )(104 m2 / cm2 )  4.48  109 W.
40. (a) Since the difference between consecutive harmonics is equal to the fundamental
frequency (see section 17-6) then f1 = (390 – 325) Hz = 65 Hz. The next harmonic after
195 Hz is therefore (195 + 65) Hz = 260 Hz.
(b) Since fn = nf1 then n = 260/65 = 4.
(c) Only odd harmonics are present in tube B so the difference between consecutive
harmonics is equal to twice the fundamental frequency in this case (consider taking
differences of Eq. 17-41 for various values of n). Therefore,
1
f1 = 2 (1320 – 1080) Hz = 120 Hz.
The next harmonic after 600 Hz is consequently [600 + 2(120)] Hz = 840 Hz.
(d) Since fn = nf1 (for n odd) then n = 840/120 = 7.