How movements influence an ultrasound
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How movements influence an ultrasound
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[post_date] => 2025-01-09 08:18:12
[post_date_gmt] => 2025-01-09 13:18:12
[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.
Ultrasound imaging is a medical technique utilized to examine inner tissue and organs of a patient. The technique is particularly useful in pregnancy imaging, as it provides a non-invasive yet thorough visualization of fetal and maternal information. This evidence then informs decisions involving dating the pregnancy, potential birth complications, and consistent monitoring of the mother’s health.
The machine first produces a sound wave, which travels into the patient. The wave’s particles are reflected or redirected in different patterns depending on the type of body tissue it comes in contact with. The sound wave then returns to the ultrasound machine, which converts the auditory information into an electrical image that is displayed on a monitor.
Scientists are interested in learning the effects of movement on ultrasonography. To study this in further detail, they place the machine on a moving apparatus. The machine emits a sound wave at the patient, who is standing a certain distance away. In order to gather further information, the scientists provide the patient with a device to measure the intensity and frequency of the waves. The new image is then compared to one in which the patient and ultrasound machine are both still.
Figure 1. A diagram of the apparatus the scientists used.
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[question] => In one trial, the machine moves towards the patient with a constant speed v. While this is occurring, which of the following descriptions is most accurate?
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[answer] => 3
[description] => Reason for the Correct Answer:
The speed of the machine is constant
Intensity gets larger as you approach the source of sound
Since the speed is constant, the Doppler shift is constant and the frequency remains constant, but as the machine approaches the patient the sound gets louder and louder since the intensity increases
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[answers] => Array
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[0] => Array
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[each_answer] => A. The frequency stays constant, and the volume stays constant
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[1] => Array
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[each_answer] => B. The wavelength gets larger and larger, and the volume stays constant
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[2] => Array
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[each_answer] => C. The frequency remains constant, and the volume continuously gets louder
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[3] => Array
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[each_answer] => D. The frequency continuously gets higher, and the volume continuously gets louder
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => In another trial, the machine speeds away from the patient with constant acceleration. What would happen to the frequency experienced by the patient?
[value] => Array
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[answer] => 1
[description] => Reason for the Correct Answer:
Think of the Doppler shift.
The source here is moving away from the observer.
Since the machine is moving away from the patient at larger and larger speeds, the frequency observed by the patient would continuously decrease
)
[answers] => Array
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[0] => Array
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[each_answer] => A. The frequency would be seen to continuously decrease
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[1] => Array
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[each_answer] => B. The frequency would be small but stay constant
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[2] => Array
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[each_answer] => C. The frequency would be seen to continuously increase
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[each_answer] => D. The frequency would be zero since sound would not make it to the patient
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => The machine is placed a location at rest and the microphone measures an intensity of 3×10-6 W/m2. Then the speaker is placed at a second location which is twice as far away from the microphone. What could be the new intensity measured by the microphone?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
Intensity is the Power divided by the area.
The area of the wave front increases like r².
Since the speaker is 2 times farther away, the intensity gets smaller by a factor of 4. So the new intensity is 0.75 W/m2.
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[0] => Array
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[each_answer] => A. 3×10-6 W/m2
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[each_answer] => B. 1.5×10-6 W/m2
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[2] => Array
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[each_answer] => C. 6×10-6 W/m2
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[3] => Array
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[each_answer] => D. 0.75×10-6 W/m2
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[quiz_unique_key] => 2261298308
[question] => During a trial the machine moves toward the patient at a speed of 30m/s and plays a frequency of 10 MHz. Assuming the speed of sound is 343m/s, what frequency is observed by the patient?
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[answer] => 3
[description] => Reason for the Correct Answer:
Use the Doppler shift formulas.
For this case the source is moving towards the observer.
The Doppler shift formula for a source moving towards the observer f’=f [343/(343-vspeaker)]
f’=440Hz [343/(343-30)]
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[answers] => Array
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[0] => Array
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[each_answer] => A. 106Hz x [343/(343+30)]
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[each_answer] => B. 106Hz x [343-30/343]
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[2] => Array
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[each_answer] => C. 106Hz x [343/(343-30)]
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[each_answer] => D. 106Hz x [343+30/343]
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[question] => At one moment the intensity of the sound at the location of the microphone is 10-5 W/m2. What is the loudness of the sound in dB at the location of the microphone (assuming the threshold of human hearing is 1012 W/m2?
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[answer] => 1
[description] => Reason for the Correct Answer:
Use the formula for decibels.
The formula for decibels is dB=10log[I/1012]
dB=10log[10-5 /10-12]=10log[107]=10 x 7 = 70dB.
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[each_answer] => A. 70dB
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[each_answer] => B. 20dB
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[2] => Array
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[each_answer] => C. 50dB
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[3] => Array
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[each_answer] => D. 60dB
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