17-5. Two loudspeakers are vibrating in phase. They are set up as in figure 17.9, and point C is located as shown there. The speed of sound is 343 m/s. The speakers play the same tone. What is the smallest frequency that will produce destructive interference?

From the figure, you can see that one of the speakers is a distance of 2.40 m from the listener, and, using Pythagoras' theorem, you can calculate that the other speaker is 4.00 m from the listener. The path difference is therefore 1.60 m.

For destructive interference, the two waves must arrive at the listener exactly out of phase, and they will do this is their path difference is a half wavelength, or one and a half wavelengths, or any odd number of half wavelengths. The smallest frequency corresponds to the longest wavelength that satisfies one of these conditions, and this is the value such that

l/2 = path difference = 1.60 m

l = 3.20 m

The frequency of a wave with this wavelength is

f = (343 m/s) / (3.20 m) = 107 Hz

17-30. A string of length 2.50 m is fixed at both ends. When the string vibrates at a frequency of 85.0 Hz, a standing wave with five loops is formed. (a) What is the wavelength of the waves that travel on the string? (b) What is the speed of the waves? (c) What would be the fundamental frequency of this string?

(a) Five loops means that there are four nodes inside the string, and that n is 5. The wavelength is (2L)/n = 2(2.50 m)/5 = 1.00 m. Alternatively, you can recognize that five loops is two and a half waves, and (2.50 m)/(2.5) = 1.00 m.

(b) v = fl = (85.0 Hz)(1.00 m) = 85.0 m/s

(c) For n = 1, l = 2L/1 = 5.00 m, and

f = v/l = (85.0 m/s)/(5.00 m) = 17.0 Hz

17-48. A person hums into the top of a well and finds that standing waves are established at frequencies of 42, 70.0, and 98 Hz. The frequency of 42 Hz is not necessarily the fundamental frequency. The speed of sound is 343 m/s. How deep is the well?

(This doesn't sound practical to me!)

Standing waves in a pipe that is closed at one end have wavelengths l = 4L, 4L/3, 4L/5, ...

The frequencies are

f = v/l = n(v/4L), n = 1, 3, 5, ...

The three frequencies 42, 70.0, and 98, must all be odd multiples of the fundamental frequency. By trial and error, you can find that they are all multiples of 14, with n = 3, 5, 7. (A more methodical approach is to see that the difference between adjacent frequencies is 2(v/4L), or twice the fundamental frequency, and the differences between 42 and 70.0, and 70.0 and 98, are both 28, so the fundamental frequency is 14 Hz. (This would not be audible, so you can't fault the whistler for not hearing it.)

The wavelength of the fundamental vibration is (343 m/s)/(14 Hz) = 24.5 m, and the depth of the well is one quarter of this, which is 6.13 m.