Section   6.2Wave Properties

1. (Exercises 3 and 4) Sound waves travel through an aluminum bar at a speed of 1400 m/s. If the frequency of the sound wave is 880 Hz, what is the distance between crests on a sound wave traveling through aluminum?

We must use the wave speed equation, which gives the relationship between wavelength, frequency, and wave speed.

v = f, or = v / f = 1400 m/s / 880 Hz = 1.6 m

Section   6.3Electromagnetic Waves

2. (Exercises 5 and 6) A beam of infrared light used to activate your TV and VCR by remote control travels at a wave speed of 3.00 x 10⁸ m/s and has a frequency of 8.25 x 10¹³ Hz. What is the wavelength of this infrared light?

v = f, or = v / f = 3.00 x 10⁸ m/s / 8.25 x 10¹³ Hz = 3.64 x 10^-6 m

Section   6.4Sound Waves

3. (Exercises 13 and 14) The ear protectors used by the workers on the taxiway at an airport are required to produce a decrease in sound intensity level from 130 dB to 70 dB or less. What is the minimum noise factor reduction produced by these ear protectors?

130 dB to 70 dB is a decrease of 60 dB, so the intensity would be reduced by a factor of 10 six times. (One factor of 10 reduction takes place for each 10 dB reduction.) This means that the reduction factor will be at least: (10 x 10 x 10 x 10 x 10 x 10) = 1.0 x 10⁶. This represents a reduction in the sound intensity being heard by the workers to one-millionth of what the intensity actually is.

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