Solutions for HW on Waves and Sound

Note on units: For all these problems it is good to bear in mind that Hz just means 1/seconds .

Ch 19.
1. a) 10 Hz b) 0.2 Hz c) 60 Hz (frequency is one over period. units are Hz = 1/sec)

2. a) .1 sec b) 5 sec c) 0.017=1/60 sec

3. If the distance between crests is 15 meters and the time between crests passing by is 5 seconds, then the waveform must be traveling 15 meters in 5 seconds, hence at a speed of 3 m/s .

4. twice each second. I think its frequency is 2 Hz and its period is 1/2=0.5 sec. The amplitude is i think 10 cm since its goes 10 above and 10 below an equilibrium point for a total excursion of 20 cm. (That is usually how amplitude is defined.)

5. wavelength is speed over frequency. In this case 3x10^8m/s divided by 100.1x10^6 Hz or about 3 meters.

7. a) period=1/f =1/(262 Hz). This is a little less that 4/1000 = .004 sec.
b) wavelength in air is 340 m/s divided by 262 Hz. This is in the neighborhood of 1.2 meters i think.

Chapter 20:
1) Wavelength of a 340Hz tone in air is about 340 m/s divided by 340 Hz = 1 meter.
Wavelength of a 34,000 Hz tone (ultrasound) is much shorter, i.e., 340 m/s divided by 34,000 Hz, which is about 1/100 =0.01 meters = 1 cm. Does that seem correct?

Chapter 21:
3) octaves correspond to factors of 2, i believe. 2000 Hz, 4000 Hz, 500 Hz, 250 Hz

5) This is a bit hard since you have to realize that for the fundamental mode the wavelength is twice the length, so in this case the wavelength is 1.5 meters. Wave speed is wavelength times frequency, which for this problem is 1.5 x 220 Hz = 330 meters/second .