I am having trouble figuring out this problem, any help would beappreciated, tha
ID: 1744259 • Letter: I
Question
I am having trouble figuring out this problem, any help would beappreciated,thanks!
Expectant parents are thrilled to hear their unborn baby'sheartbeat, revealed by an ultrasonic motion detector. Suppose thefetus's ventricular wall moves in simple harmonic motion with anamplitude of 1.70 mm and a frequencyof 140 per minute.
(a) Find the maximum linear speed of the heartwall. Suppose the motion detector in contact with the mother'sabdomen produces sound at 2,000,000.0 Hz which travels throughtissue at 1.50 km/s.
1 m/s
(b) Find the difference between the original frequency and themaximum frequency at which sound arrives at the wall of the baby'sheart.
2 Hz
(c) Find the difference between the original frequency and themaximum frequency at which reflected sound is received by themotion detector. By electronically "listening" for echoes at afrequency different from the broadcast frequency, the motiondetector can produce beeps of audible sound in synchronization withthe fetal heartbeat.
3 Hz (a) Find the maximum linear speed of the heartwall. Suppose the motion detector in contact with the mother'sabdomen produces sound at 2,000,000.0 Hz which travels throughtissue at 1.50 km/s.
1 m/s
(b) Find the difference between the original frequency and themaximum frequency at which sound arrives at the wall of the baby'sheart.
2 Hz
(c) Find the difference between the original frequency and themaximum frequency at which reflected sound is received by themotion detector. By electronically "listening" for echoes at afrequency different from the broadcast frequency, the motiondetector can produce beeps of audible sound in synchronization withthe fetal heartbeat.
3 Hz
Explanation / Answer
(a) from the theory we know that = 2 f = 2 (105 / min /60.0 s / min) = ....... rad /s the maximum linear speed will be vmax = A =() (2.00 x 10-3 m) =........ m / s (b) from the given we can see that the heartwall is a moving observer and the detector is astationar source so we get f ' = f ((v + vmax) / v) = (2 x 106 Hz)[(1500 + vmax) / (1500)] = ......... Hz (c) in this case the heart wall is a moving sourceand the detector will be a stationary source f '' = f ' (v / (v- vmax)) = ......... Hz amplitude a = 2.00 mm = ........m frequency f = 105 per min = ........ Hz frequency (f ') of source of sound = 2MHz = 2 * 106 Hz velocity = 1.50 * 103m/s (a) first we find the angular frequency = 2 f = ........rad /s for harmonic motion we get the maximumlinear speed of the heart wall as vmax = a = ........... m/s (b) from the doppler effwct we can write f ' = (v + vo / v - vs)f where vo = vmax is thevelocity of the observer and vs is the velocity of thesource of sound if heart wall is assumed as the observer it willbe moving and the detector (source of sounnd) is said to be stationary that meansvs = 0, then by applying the sign convention to thedoppler effect according to the given condition we get f ' = (v + vmax / v) = .........HzRelated Questions
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