Question

An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta leading away from the heart. Since blood within the heart is essentially stationary, this pressure difference can be inferred from a measurement of the speed of blood flow in the aorta. Take the speed of sound in stationary blood to be c.

a. Sound sent by a transmitter placed directly inline with the aorta will be reflected back to a receiver and show a frequency shift with each heartbeat. If the maximum speed of blood in the aorta is v, what frequency will the receiver detect? Note that you cannot simply use the textbook Doppler Shift formula because the detector is the same device as the source, receiving sound after reflection.
b. Show that in the limit of low blood velocity (v <
f= 2fo v/c

Answer 1

a) f ’’ = f₀
`(1 + (v)/(c) )/(1- (v)/(c) )` , b) Δf = 2 f₀
`(v)/(c)`

Step-by-step explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

f ’= fo
`(c+v)/(c)`

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

f ’’ = f’
`(c)/( c-v)`

where c represents the sound velocity in stationary blood

therefore the received frequency is

f ’’ = f₀
`(c)/(c-v)`

let's simplify the expression

f ’’ = f₀ \frac{c+v}{c-v}

f ’’ = f₀
`(1 + (v)/(c) )/(1- (v)/(c) )`

b) At the low speed limit v <c, we can expand the quantity

(1 -x)ⁿ = 1 - x + n (n-1) x² + ...

`( 1- (v)/(c) ) ^(-1) = 1 + (v)/(c)`

f ’’ = fo
`( 1+ (v)/(c)) ( 1 + (v)/(c) )`

f ’’ = fo
`( 1 + 2 (v)/(c) + (v^2)/( c^2) )`

leave the linear term

f ’’ = f₀ + f₀ 2
`(v)/(c)`

the sound difference

f ’’ -f₀ = 2f₀ v/c

Δf = 2 f₀
`(v)/(c)`