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# Waves Study Guide (page 3)

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#### Example

The laboratory motion sensors used in mechanics experiments are ultrasonic sensors that send a wave out and measure the amount of time to its return. By knowing the time and the speed of the sound, you can measure the position of different objects relative to the position of the sensor. Considering the speed of sound to be 343 m/s in a room where the temperature is 20° C, find the mass of a molecule of the gas present in the room. (γ is 1.2.)

#### Solution

First, inventory your known data and then set the equation necessary to find the mass.

v = 343 m/s

T = t + 273.15 K = 20° C + 273.15 K = 293.15 K

γ = 1.2

m = ?

v = =

v =

v2 =

m =

m =

m = 1.41 · 10–23 kg

In the case of a liquid, the speed of sound is:

v =

where K is a coefficient that depends on the pressure change with volume and ρ is the density of the liquid. Both factors are material dependent.

And lastly, the speed of sound is also affected by the nature of the solid in which it travels. The expression is:

v =

where E is Young's modulus that measures the elastic properties of an object subject to a force, and ρ is the density of the solid.

## Electromagnetic Waves

Mechanical sources produce mechanical waves that require a medium in which to propagate, but there are other sources that produce waves that can travel through a vacuum. The fluctuation of a magnetic and electric field can produce a propagating wave called an electromagnetic wave.

How is it possible to have propagation without mass? Remember a few lessons ago, we learned that a current can induce a magnetic field and a magnetic field can induce a current. The two factors together can propagate through space with no requirement of matter. The magnetic and electric field fluctuation was studied by James Maxwell (1831-1879), and the diagram of the effect is shown in Figure 17.8.

Although electromagnetic waves propagate through media, the maximum speed is achieved in a vacuum. And for the purpose of this lesson, we will work with a value of:

c = 3.0 · 108 m/s

The electromagnetic spectrum contains the visible spectrum, or light as we call it. But this is only a small part of the range of electromagnetic frequencies, which vary from 4.05 · 1014 to 7.9 · 1014 Hz. Other components of the electromagnetic spectrum are listed in Table 17.l.

The wavelength for the ranges of frequency can be easily found by using the definition of the speed of the wave:

v = f · λ

λ =

For a vacuum, the speed of an electromagnetic wave is c:

λ =

#### Example 1

Find the range of wavelength for visible light if the light spectrum corresponds to the range from 4.05 · 1014 to 7.9 · 1014Hz.

#### Solution 1

Set up the previous equation with the proper data and solve for the frequency.

λ = = = 740.7 · 10–9m

A unit very often used in optics is the nanometer: 1 nm = 10–9 m.

λ = = = 379.7 · 10–9m

Hence, we have a range of 380 nm to 741 nm.

The energy carried by the electromagnetic wave is dependent on the properties of the two propagating fields, and its expression is:

u = · ε0 · E2 + · B2

where E and B are the electric and magnetic fields, and ε0 and μ0 are the permittivity and permeability of vacuum.

ε0 = 8.85 · 10–12 C2/N · m2 and μ = 4 · π · 10 –7 T · m/ A

In a vacuum, the electric and magnetic fields are related by:

E = c · B

And the speed in vacuum is given by:

c =

#### Example 2

Reduce the expression of the energy of the electromagnetic waves to a simpler form using the connection between the electric field and the magnetic field.

#### Solution 2

Start with the definition of the energy, and, using the previous expression, try to simplify it.

u = · ε0 · E2 + · B2

u = · ε0 · (c · B)2 + · B2

u = · B2

Practice problems of this concept can be found at: Waves Practice Questions

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