Question

A red blood cell travels at speed 0.40 m/s in a large artery. A sound frequency 2.00*10^6 Hz enters the blood opposite the direction of flow. a) Determine the frequency of sound reflected from the cell and detected by a reciever. b) If the emitted and received sounds are combined in the receiver, what beat frequnecy is measured?

Answer 1

Solution)

Given,

frequency of sound n=2*10^6 Hz

We know, velocity of sound V=1500 m/sec

Given, Velocity of the blood cell Vb=0.4 m/sec

So,

Frequency at blood cell is n1=n*[(V+Vb)/V]

n1=2*10^6*[(1500+0.4)/1500]

So, n1=2.00053*10^6 Hz

now, blood cell acts as source of sound and,

frequency at receiver is n2=n1*[V/(V-Vb)] (here, source is moving towards the observer)

n2=2.000533*10^6*[1500/(1500-0.4)]

n2=2.001067*10^6 Hz

n2=2.001067 Hz (Ans)

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B) n_beat=n2-n

=2.001067*10^6-2*10^6

=1067 Hz (Ans)

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Good luck!:)