Resonance in a tube
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- Resonance in a tube
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Resonance in a tube
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[post_date] => 2025-01-09 08:19:46
[post_date_gmt] => 2025-01-09 13:19:46
[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
When an external source plays a frequency that matches one of the harmonic standing waves of an object, a resonance can occur within the object. Resonance involves the reinforcement of a wave due to synchronized reflection of the wave. These “perfect reflections” can be induced by manipulating the neighboring environment, and are often used in medicine, such as in Magnetic Resonance Imaging (MRI). Resonance is exemplified by a large amount of energy transfer to the object and a dramatic increase in amplitude.
To investigate resonance, an experiment is performed wherein a speaker plays sound towards a tube of length D=2.0 meters which is open at both ends. The frequency of the speaker can be adjusted. It is observed that for certain speaker frequencies standing sound waves are set up in the tube causing a large resonance. Assume the speed of sound is 343m/s.
Figure 1. The apparatus of the experiment. The speaker produces sound waves which travel through the tube.
[post_title] => Resonance in a tube
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[question] => What is the fundamental wavelength that can be set up for this tube?
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[answer] => 4
[description] => Reason for the Correct Answer:
This is an open-open tube which will have displacement antinodes at both ends.
The fundamental wavelength is the largest possible wavelength standing wave in the tube.
Remember that λ=2L/n.
Remember that λ=2L/n=2(2.0)/1=4.0m).
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[answers] => Array
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[0] => Array
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[each_answer] => A. 1.0m
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[1] => Array
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[each_answer] => B. 2.0m
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[2] => Array
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[each_answer] => C. 3.0m
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[3] => Array
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[each_answer] => D. 4.0m
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => What frequency could the speaker play to set up the fourth harmonic standing wave for this tube?
[value] => Array
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[answer] => 1
[description] => Reason for the Correct Answer:
First find the wavelength of the fourth harmonic.
λ=2L/n=2(2.0)/4=1.0m.
Remember that v=λf.
v=λf so f=v/λ=(343m/s)/(1.0m)= 343Hz.
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[answers] => Array
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[0] => Array
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[each_answer] => A. 343 Hz
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[1] => Array
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[each_answer] => B. 343/2 Hz
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[2] => Array
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[each_answer] => C. 343 x 2 Hz
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[3] => Array
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[each_answer] => D. 343/4 Hz
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => You find that when the speaker plays a frequency of 686Hz the tube resonates loudly. Which harmonic corresponds to the frequency of 686Hz? (Assume the first harmonic n=1 is the fundamental frequency, the second harmonic is n=2, and so on.)
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[answer] => 3
[description] => Reason for the Correct Answer:
Figure out which harmonic corresponds to 686Hz.
Since the frequency is 686Hz, the wavelength is λ=v/f=0.5m).
Now use the formula for λ=2L/n and solve for n which gives n=2L/ λ=2(2.0m)/0.5m=8.
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[answers] => Array
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[0] => Array
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[each_answer] => A. 4
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[1] => Array
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[each_answer] => B. 6
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[2] => Array
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[each_answer] => C. 8
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[3] => Array
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[each_answer] => D. 10
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[3] => Array
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[quiz_unique_key] => 2261298308
[question] => If one were to increase the temperature of the air in the tube, how would that affect the fundamental resonant frequency observed in the tube?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
The speed of sound depends on the temperature of the medium.
The speed of sound increases in hotter air.
The fundamental wavelength would stay the same, since the length of the tube is not changing.
Since f=v/λ and v increases and λ stays the same, the fundamental frequency would increase.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. The fundamental resonant frequency in the tube would not change, since the length of the tube is not changing
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[1] => Array
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[each_answer] => B. The fundamental resonant frequency in the tube would decrease, since the fundamental wavelength would decrease
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[2] => Array
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[each_answer] => C. The fundamental resonant frequency in the tube would increase, since the speed of sound would increase in hotter air106Hz x [343/(343-30)]
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[3] => Array
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[each_answer] => D. The fundamental resonant frequency in the tube would decrease, since the sound takes longer to travel down the tube in hotter air
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[quiz_unique_key] => 2377279144
[question] => If the bottom of the tube were made to be closed instead of open, how would that affect the fundamental wavelength set up in the tube?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
Now the tube is open-closed with a node at one end and an antinode at the other end.
The formula for the wavelengths in open closed tubes is λ=4L/n.
Since the wavelength of the fundamental is now 4L instead of 2L the fundamental wavelength is doubled by closing the bottom of the tube.
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[answers] => Array
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[0] => Array
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[each_answer] => A. The fundamental wavelength would not change70dB
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[each_answer] => B. The fundamental wavelength would quadruple
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[2] => Array
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[each_answer] => C. The fundamental wavelength would be cut in half
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[each_answer] => D. The fundamental wavelength would double
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