Name________________KEY__________________________________ Date______11/14/14____________
Investigation 10
Waves and Sound
1.
The two diagrams show waves on a string that are traveling toward each other. Draw what happens
when the waves meet and what happens a short time after they meet.
a.
b.
c.
2.
A source of sound (e.g., a siren, a speaker, a tuning fork, etc.) is emitting sound in all directions. The
frequency of the sound is constant. Draw a diagram showing the wavefronts being emitted from the
sound source. (Show at least five wavefronts.)
a.
The source of sound is stationary.
b.
The source of sound is moving to the right.
c.
What does the distance from one wavefront to the next wavefront represent?
The distance from wavefront to the next equals the wavelength of the waves.
d.
When the source of sound is moving to the right, what happens to the
1)
wavelength of the sound?
The wavelength decreases to the right of the source and increases to the left of the source.
2)
frequency of the sound?
The frequency increases to the right of the source and decreases to the left of the source.
3.
A harmonic, longitudinal wave is sent along a slinky. The distance from a rarefaction to an adjacent
compression is 0.5 meters. The waves are traveling along the slinky at a speed of 2 m/sec.
a.
What is the value of the wavelength of the wave?
Wavelength is from one rarefaction to the next rarefaction = 0.5 +0.5 = 1.0 meters.
b.
What is the value of the frequency of the wave?
Frequency = (speed of the wave)/(wavelength of the wave) = (2 m/s)/(1 meter) = 2 Hertz.
4.
Sound waves travel faster through steel than air. Explain why.
Sound waves travel as a result of collisions between adjacent molecules. The molecules are closer
together in steel than in air, and the collisions take a lot less timed to occur.
5.
The fundamental frequency of a 1.2 meters long guitar string
is 220 Hz. The shape of the oscillations on the string when it
is oscillating in its fundamental frequency is shown.
a.
What is the wavelength of the wave on the string?
wavelength = 2 x 1.2 m =2.4 m
b.
How fast are the waves traveling on the string?
Speed = frequency x wavelength = (220 Hz) x (2.4 m) = 528 m/s
The string is plucked again and is then lightly touched at a
point that is one-third the length of the string away from
the end of the string producing three loops. (See diagram.)
c.
How many nodes are there?
4 nodes
d.
How many antinodes are there?
3 antinodes
e.
What is the name of this harmonic?
3rd harmonic
f.
What is the name of this overtone?
2nd overtone
g.
What is the wavelength of the standing waves?
Wavelength = (2/3) x (1.2 m) = 0.8 m
h.
What is the frequency of oscillation of the string?
Frequency = speed/wavelength = (528 m/s)/(0.8 m) = 660 Hz,
or 3 x fundamental = 3 x 220 Hz = 660 Hz
1.2 m
6.
The air in a tube that is open at both ends is resonating in its fundamental frequency (first harmonic).
The speed of sound is 345 m/s and the length of the tube is 0.20 m.
a.
What is the wavelength of the wave in the tube?
Wavelength = 2 x 0.20 m = 0.40 m
b.
Find the frequency at which the air in the tube is resonating.
Frequency = speed/wavelength = (345 m/s)/(0.40 m) = 862.5 Hz
c.
If the air in the tube is resonating in its first overtone, then find that frequency.
First overtone = second harmonic = 2 x fundamental = 2 x 862.5 = 1725 Hz
7.
Suppose the tube is open at one end and closed at the other, and is oscillating (or resonating) in its
fundamental frequency (first harmonic) of 328 Hz. The distance from the open end of the tube to the
closed end of the tube is 0.26 meters.
a.
The closed end of the tube acts like a node of the oscillation. What is a node?
The particles at the nodes do not oscillate.
b.
The open end of the tube acts like an antinode of the oscillation. What is an antinode?
The particles undergo maximum oscillation.
c.
Nodes and antinodes alternate in a standing wave. The diagram below shows alternating
antinodes (A) and nodes (N). Starting with the antinode on the left, indicate how long one
wavelength is. Also, starting with the node on the far right, indicate how long one wavelength is.
A
d.
N
A
N
A
N
A
N
A
N
What is the value of the wavelength of the sound wave that is resonating in the 0.26 meter long
open-closed tube?
Wavelength = 4 x 0.26 m = 1.04 m
e.
What is the value of the speed of sound in the air in the tube?
Speed = frequency x wavelength = (328 Hz) x (1.04 m) = 341 m/s