11. The trumpet player now pushes both the 2nd and 3rd valves, opening both bits of extra tubing. What is the new length L of the horn? What is the frequency of the 6th harmonic produced?

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Trumpet Physics
As we learned in the video lesson, an open pipe of length L can create standing
waves at certain specific frequencies, given by:
fn
= n f1
2L
= n
fn = the frequency that creates a standing wave with n loops, in Hz
= the "nth harmonic"
fi = the frequency that creates a standing wave with 1 loop, in Hz
= the "fundamental frequency"
n = number of loops in the standing wave
v = speed of the sound wave = 343 m/s (at 20° C)
L= length of the pipe, in m
%3D
The fundamental frequency, f1, is the lowest frequency that makes a standing wave on
the pipe. The higher frequencies or “harmonics" are multiples of the fundamental: f2
is twice as much as f1; f3 is three times as big as f1, and so on.
A trumpet is essentially a long, curved open pipe. (Actually, its geometry is
more complicated, but it is designed to produce the same set of frequencies that an
open pipe would. So close enough.) It has three valves, each of which is connected to
an extra bit of tubing.
When no valves are pushed, the sound wave passes straight through the horn,
travelling a distance L. When a valve is pushed, it opens up the extra bit of tubing,
which increases the length L of the horn slightly. According to the equation above,
increasing the length L lowers the frequency of the standing wave, which produces a
lower tone or note. So, when a valve is pushed, it lowers the note the horn produces.
Pushing a valve with a longer tube lowers the note even more. Opening more than one
valve makes the sound travel through more than one extra bit of tubing, lowering the
note even further. By picking the right combination of valves, the trumpet can make
any note on the musical scale.
Directions: Here are some practice problems to make you familiar with these
concepts. Please review the following. Assume the speed of sound is 343 m/s.
Transcribed Image Text:Trumpet Physics As we learned in the video lesson, an open pipe of length L can create standing waves at certain specific frequencies, given by: fn = n f1 2L = n fn = the frequency that creates a standing wave with n loops, in Hz = the "nth harmonic" fi = the frequency that creates a standing wave with 1 loop, in Hz = the "fundamental frequency" n = number of loops in the standing wave v = speed of the sound wave = 343 m/s (at 20° C) L= length of the pipe, in m %3D The fundamental frequency, f1, is the lowest frequency that makes a standing wave on the pipe. The higher frequencies or “harmonics" are multiples of the fundamental: f2 is twice as much as f1; f3 is three times as big as f1, and so on. A trumpet is essentially a long, curved open pipe. (Actually, its geometry is more complicated, but it is designed to produce the same set of frequencies that an open pipe would. So close enough.) It has three valves, each of which is connected to an extra bit of tubing. When no valves are pushed, the sound wave passes straight through the horn, travelling a distance L. When a valve is pushed, it opens up the extra bit of tubing, which increases the length L of the horn slightly. According to the equation above, increasing the length L lowers the frequency of the standing wave, which produces a lower tone or note. So, when a valve is pushed, it lowers the note the horn produces. Pushing a valve with a longer tube lowers the note even more. Opening more than one valve makes the sound travel through more than one extra bit of tubing, lowering the note even further. By picking the right combination of valves, the trumpet can make any note on the musical scale. Directions: Here are some practice problems to make you familiar with these concepts. Please review the following. Assume the speed of sound is 343 m/s.
2. When the first valve is pushed, the frequency of the sixth harmonic changes to
fo = 622 Hz (an E-flat). Calculate the new length of the pipe L.
3. The new length is slightly longer than the original length, because of the added
length of the tubing attached to the first valve. Subtract the two lengths to find
the length of the tube attached to the first valve.
4. When the second valve (only) is pushed, the frequency of the 6™ harmonic changes
to fo = 659 Hz (an E). What is the new length of the pipe L?
5. The difference in length between Step 4 and Step 1 is due to the length of the tubing
attached to the second valve. Calculate the length of the tube attached to the
second valve.
6. When the third valve (only) is pushed, the frequency of the 6™ harmonic changes to
fo = 585 Hz (a D). What is the new length of the pipe L?
7. The difference in length between Step 6 and Step 1 is due to the length of the tubing
attached to the third valve. Calculate the length of the tube attached to the third
valve.
8. The trumpet player now pushes both the 1st and 2™d valves, opening both bits of
extra tubing. What is the new length L of the horn?
9. What is the frequency of the 6™ harmonic produced in Problem 8?
10. The trumpet player now pushes both the 1d and 3d valves, opening both bits of
extra tubing. What is the new length L of the horn? What is the frequency of
the 6th harmonic produced?
11. The trumpet player now pushes both the 2nd and 3rd valves, opening both bits of
extra tubing. What is the new length L of the horn? What is the frequency of
the 6th harmonic produced?
12. The trumpet player now pushes all three of the valves, opening all three bits of
extra tubing. What is the new length L of the horn? What is the frequency of
the 6th harmonic produced?
13. For the 4" harmonic, what is frequency produced when the 1ª and 3d valves are
pushed?
Transcribed Image Text:2. When the first valve is pushed, the frequency of the sixth harmonic changes to fo = 622 Hz (an E-flat). Calculate the new length of the pipe L. 3. The new length is slightly longer than the original length, because of the added length of the tubing attached to the first valve. Subtract the two lengths to find the length of the tube attached to the first valve. 4. When the second valve (only) is pushed, the frequency of the 6™ harmonic changes to fo = 659 Hz (an E). What is the new length of the pipe L? 5. The difference in length between Step 4 and Step 1 is due to the length of the tubing attached to the second valve. Calculate the length of the tube attached to the second valve. 6. When the third valve (only) is pushed, the frequency of the 6™ harmonic changes to fo = 585 Hz (a D). What is the new length of the pipe L? 7. The difference in length between Step 6 and Step 1 is due to the length of the tubing attached to the third valve. Calculate the length of the tube attached to the third valve. 8. The trumpet player now pushes both the 1st and 2™d valves, opening both bits of extra tubing. What is the new length L of the horn? 9. What is the frequency of the 6™ harmonic produced in Problem 8? 10. The trumpet player now pushes both the 1d and 3d valves, opening both bits of extra tubing. What is the new length L of the horn? What is the frequency of the 6th harmonic produced? 11. The trumpet player now pushes both the 2nd and 3rd valves, opening both bits of extra tubing. What is the new length L of the horn? What is the frequency of the 6th harmonic produced? 12. The trumpet player now pushes all three of the valves, opening all three bits of extra tubing. What is the new length L of the horn? What is the frequency of the 6th harmonic produced? 13. For the 4" harmonic, what is frequency produced when the 1ª and 3d valves are pushed?
Expert Solution
Step 1

11) The length of the trumpet pipe is found by :

L=n2fnv

When there is no valve pushed, the length of the pipe is :

n=6, f6=698Hz, v=343m/sL=62×698×343=1.474m

When the second valve is pushed, the length of the pipe is :

n=6, f6=659Hz, v=343m/sL2=62×659×343=1.56m

When the third valve is pushed, the length of the pipe is :

n=6, f6=585Hz, v=343m/sL3=62×585×343=1.759m

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