1502 Periodic Waves

The wavelength λ\lambda of the wave is the distance from one crest to the next, or from one trough to the next, or from any poiint ot the corresponding point on the next repetition of the wave shape. The wave pattern travels with constant speed vv and advances a distance of one wavelength λ\lambda in a time interval of one period TT. So the wave speed is v=λ/Tv = \lambda /T, or because f=1/Tf = 1/T.

v=λf\begin{aligned} v = \lambda f \end{aligned}

Exercises

1, 2, 4, 5

15.1 The speed of sound in air at 20/degree20/degree is 344 m/s. (a) What is the wavelength of a sound wave with a frequency of 784 Hz, corresponding to the note G5G_5 on a piano, and how many milliseconds does each vibration take? (b) What is the wavelength of a sound wave one octave higher (twice the frequency) than the note in part (a)?

Solution

a. Wavelength and Period

fa=784Hzλa=vfa=344784=0.439mT=1fa=1784×1000=1.28ms\begin{aligned} f_a &= 784 Hz\\ \lambda_a &= \frac{v}{f_a} = \frac{344}{784} = 0.439m\\ T &= \frac{1}{f_a} = \frac{1}{784} \times 1000 = 1.28ms \end{aligned}

b. Wavelength of a sound save with twice the frequency in part a.

λb=vfb=344784×2=0.219m\begin{aligned} \lambda_b &= \frac{v}{f_b} = \frac{344}{784 \times 2} = 0.219m\\ \end{aligned}

15.2 BIO Audible Sound. Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from about 20.020.0 Hz to about 20.020.0 kHz. (a) If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound and a blue dot for the short-wavelength sound, how far apart would the red dots be, and how far apart would the blue dots be? (b) In reality would adjacent dots in each set be far enough apart for you to easily measure their separation with a meter stick? (c) Suppose you repeated part (a) in water, where sound travels at 14801480 m/s. How far apart would the dots be in each set? Could you readily measure their separation with a meter stick?

Solution

Todo

15.4 BIO Ultrasound Imaging. Sound having frequencies above the range of human hearing (about 20,000 Hz) is called ultrasound. Waves above this frequency can be used to penetrate the body and to produce images by reflecting from surfaces. In a typical ultrasound scan, the waves travel through body tissue with a speed of 15001500 m/s. For a good, detailed image, the wavelength should be no more than 1.0 mm. What frequency sound is required for a good scan?

Solution

λ=10−3mv=1500m/sf=vλ=150010−3=1.5×106Hz\begin{aligned} \lambda &= 10^{-3} m\\ v &= 1500m/s\\ f &= \frac{v}{\lambda} = \frac{1500}{10^{-3}} = 1.5 \times 10^6 Hz \end{aligned}

15.5 BIO (a) Audible wavelengths. The range of audible frequencies is from about 20 Hz to 20,000 Hz. What is the range of the wavelengths of audible sound in air? (b) Visible light. The range of visible light extends from 380 nm to 750 nm. What is the range of visible frequencies of light? (c) Brain surgery. Surgeons can remove brain tumors by using a cavitron ultrasonic surgical aspirator, which produces sound waves of frequency 23 kHz. What is the wavelength of these waves in air? (d) Sound in the body. What would be the wavelength of the sound in part (c) in bodily fluids in which the speed of sound is 14801480 m/s but the frequency is unchanged?

Solution

a. Audible wavelength

λlo=vsfhi=34420000=0.0172mλhi=vsflo=34420=17.2m\begin{aligned} \lambda_{lo} &= \frac{v_s}{f_{hi}} = \frac{344}{20000} = 0.0172m\\ \lambda_{hi} &= \frac{v_s}{f_{lo}} = \frac{344}{20} = 17.2m \end{aligned}

b. Light

λlo=vlfhi=3.0×108750×10−6=4×1011mλhi=vlflo=3.0×108380×10−6=7.89×1011m\begin{aligned} \lambda_{lo} &= \frac{v_l}{f_{hi}} = \frac{3.0 \times 10^8}{750 \times 10^{-6}} = 4 \times 10^{11}m\\ \lambda_{hi} &= \frac{v_l}{f_{lo}} = \frac{3.0 \times 10^8}{380 \times 10^{-6}} = 7.89 \times 10^{11}m \end{aligned}