Expecting parents are thrilled to hear their unborn baby\'sheartbeat, revealed b

Question

Expecting 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 withamplitude 1.80mm and frequency 115 per minute.
a. find the maximum linear speed of the heart wall.
Suppose the motion detector in contact with the mothers abdomenproduces sound at percisely 2MHz, which travel through tissue at1.5 km/s
b. find the maximum frequency at which sound arrives at the wall ofthe baby's heart.
c. find the maximum frequency at which reflected sound is receivedby the motion detector.
(by electronically "listening for echoes at a frequency differentfrom the broadcast frequency, the motion detector can produce beepsof audible sound is synchrony with the fetal heartbeat.)
Expecting 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 withamplitude 1.80mm and frequency 115 per minute.
a. find the maximum linear speed of the heart wall.
Suppose the motion detector in contact with the mothers abdomenproduces sound at percisely 2MHz, which travel through tissue at1.5 km/s
b. find the maximum frequency at which sound arrives at the wall ofthe baby's heart.
c. find the maximum frequency at which reflected sound is receivedby the motion detector.
(by electronically "listening for echoes at a frequency differentfrom the broadcast frequency, the motion detector can produce beepsof audible sound is synchrony with the fetal heartbeat.)

Explanation / Answer

(a)    from the theory we know that     = 2 f        = 2 (115 / min /60.0 s / min)           = ....... rad /s    the maximum linear speed will be    vmax = A           =() (1.80 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 = 1.80 mm                      = ........m    frequency f = 110 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)        = .........Hz

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