Let's begin our analysis of wave motion by considering the properties of a sound wave.

1. A tuning fork produces a sound wave in air having a wavelength of 1.30 meters. At what frequency does this tuning fork vibrate?

Here we can use the wave speed equation, where the wave travels at the speed of sound through air. There is a slight temperature and pressure dependence for the velocity of sound in air, so we will assume room temperature and normal atmospheric pressure and use the value for the speed of sound that is given in the textbook, 344 m/s.

v = f, or f = v / = 344 m/s / 1.30 m = 265 Hz

Now we will consider a typical example of how a standing wave can be set up and used to produce sound waves in the air, such as when musical notes are established in resonant air columns or on stretched strings.

2. What will be the wave speed for the wave traveling along the violin string playing at a frequency of 392 hertz and having a wavelength of 0.760 m?

Since we know both the frequency and the wavelength of the wave, we can calculate the wave speed on the string.

v = f = 0.760 m (392 Hz) = 298 m/s

Note that this is not the wave speed for sound waves produced in the air by the vibrating string but the actual speed at which the wave energy moves back and forth along the string itself.

Electromagnetic waves such as light and radio waves also follow the rules we have been using to solve the previous exercises, but our analysis of these waves is a little more difficult because the numbers needed to describe their wave speed, frequency, and wavelength are sometimes so large or so small that they require the use of powers-of-ten notation. We will also show the calculations as they would be done in conventional notation. This may also help you to understand why powers-of-ten notation is often required, even for calculations of a quite simple nature.

3. A Cincinnati radio station broadcasts on a frequency of 700 kilohertz. What is the wavelength of its transmitted radio waves?

We must be careful here because, although we listen to the sounds produced by a radio broadcast, the signals sent out by the station are really electromagnetic waves that travel at the speed of light. We must have a receiver, an amplifier, and a speaker that can convert these radio waves into sound waves before our ears can detect them. The wave velocity is then: v = c = 3.00 x 10⁸ m/s.

The wavelength of this radio signal is found as follows.

c = f, or = c / f = 3.00 x 10⁸ m/s / 700 x 10³ Hz

or in conventional notation: = 300,000,000 m/s / 700,000 Hz

= 4.29 x 10² m or 429 m

This is well over the length of four football fields. Radio signals have very long wavelengths. Did you remember to convert 700 kilohertz to hertz before you did the above calculation? "Kilo" means 1000, so the correct value may be expressed as 7.00 x 10⁵ Hz, 700 x 10³ Hz, or 700,000 Hz. As you can see, in the above example we left the 700 and simply put in the 1000 using powers-of-ten notation (10³).

Calculations like this can be done for other types of waves as well. Consider Example 6.1 in the textbook, where this same method of calculation is used to find the wavelength of sound waves in air.

Returning to waves that we can actually see as they propagate from one place to another, we can look at another example using the disturbances moving across the surface of a still body of water.

4. Waves moving on a lake are observed to have a speed of 2.0 m/s, and to have a distance of 5.0 m between wave crests.

a) Determine the frequency of the waves.
b) Find the period of the wave motion.

The wave speed is 2.0 m/s and the distance between wave crests, 5.0 m, is the wavelength. We can then use the wave speed equation.

a)v = f, or f = v / = 2.0 m/s / 5.0 m = 0.40 1/s = 0.40 Hz

Remember that the period is simply the reciprocal of the frequency so:

b)T = 1 / f = 1 / 0.40 Hz = 2.5 s

This means that a boat floating on the surface of the lake would bob up and down once every 2.5 seconds. Let's hope the amplitude of these waves is not very high, or the people in the boat will all be seasick!

Now let's consider how the intensity of a sound wave varies, using the units of decibels.

5. A speaker is playing with an intensity level of 80 dB. If the volume is turned up so that the intensity is 10,000 times greater, what will be the new intensity level?

Each increase in intensity of a factor of 10 is an increase of 10 dB, so for an increase of 10,000 (10 x 10 x 10 x 10) there should be a (4 x 10), or a 40 dB increase. The new intensity level is then just 80 dB + 40 dB = 120 dB.

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