Spring 2005
Dr. Mike Fanelli
Solutions to Assigned Problems
Chapter 3
PROBLEM 3-1:
A sound wave moving through water has a frequency of 256 Hz (hertz or wave
cycles per second) and a wavelength of 5.77 meters. What is the speed of
sound in water ?
ANSWER: The general expression relating wavelength, frequency, and the
speed of any wave is:
 (wavelength)
x
 (frequency)
= speed of the wave,
where wavelength is expressed in linear unit (meters, kilometers, miles, for
example), and frequency is expressed in cycles per second (a unit also known as
a hertz).
5.77 meters x 256 Hz = 1480 meters / second.
This value is a little less than 1 mile per second.
PROBLEM 3-2:
What is the wavelength of a 100-Mhz (megahertz) radio signal, equivalent to the
wavelength of your FM radio tuned to 100 on the dial ?
ANSWER: Radio waves are a form of electromagnetic radiation, unlike sound
waves, which are a type of density wave. Radio waves, like all EM waves, travel
at the same speed: the “speed of light”. As in problem 1,
 (wavelength) x
 (frequency) = speed of wave,
You are given the frequency, 100 MHz = 100 million hertz, and the speed of light
= 300,000 kilometers per second = 300,000,000 meters per second (since 1 km
= 1000 meters). Dividing each side of the expression by the frequency gives:
 (wavelength)
=
300,000,000 m/s
----------------------- = 3 meters
100,000,000 hertz
PROBLEM 3-7:
Your normal body temperature is about 37 C. Express this temperature using
the Kelvin temperature scale, then determine the peak wavelength of your
“emission”, assuming the human body acts like a blackbody.
ANSWER: Using MP 3-1, which describe the three temperature scales:
Kelvins = degrees Celsius + 273. Therefore, 37 + 273 = 310 K
The typical temperature of humans (and other objects on Earth’s surface) is
around 300 K
Using MP 3-2, which discusses radiation laws, we apply Wien’s law.
0.29
(max) = ----------, expressed in centimeters
T
Given that T = 310 K, as determined above, (max) = 9.35 x 10-4 centimeters =
9.35 microns, since 1 micron = 10-4 centimeters.
Using Figure 3.3 in the text, one can see that this wavelength is located in the
infrared part of the spectrum. People can be seen at night, i.e., in the absence of
ambient visible light, using “thermal” imaging systems, devices that can detect
infrared light at wavelengths of 8-10 microns.
PROBLEM 3-10:
You are given two objects, radiating like blackbodies, at temperatures of 300 K
and 1500 K. Which is producing more energy and by how much ?
ANSWER: Comparison of the total energy (radiation) output from objects is an
application of Stefan’s “Law”. Recall that:
4
Total energy emission is proportional to temperature
Interpreting Stefan’s Law:


Hotter objects emit more energy
The ratio of energy emitted between two objects will depend on their
temperature ratio to the fourth power.
4
4
Therefore, the 1500 K object radiates more energy, by (1500  300)
= 5 =
625 times more energy.