Question

How the doppler effect is used in medicine?

Answer 1

In medicine, the Doppler Effect can be used to measure the direction and speed of blood flow in arteries and veins.

Answer 2

Explanation:

DOPPLER'S EFFECT

Non - Relativistic Treatment

As most readers must have read in physics according to. This phenomenon the pitch sound observed by an observer is different in two cases, firstly when both the source and observer are stationary and secondly when there is relative motion between them. When source and observer are approaching each other, the apparent frequency is increased and when receding, decreased. This phenomenon is called doppler's effect which occurs with all kinds of wave motion, although in case of mechanical waves involving a material medium, its nature is some what different from the case of electromagnetic waves such as light where no medium is involved . In the case of light waves, if spectrum is observed in a spectrometer, a shift in the position of spectral line from the original position is observed with the motion of source.

we first consider the case when the observer is at rest and the source is moving towards it velocity u. let the frequency of the source be f so that the time period

T=1/f

and the wave speed be c in a time time interval T during which one cycle of wave is emitted the wave progresses a at distance cT but in the same period the source wave in the same direction distance u T hence wavelength which is distance between two successive maximum in the wave is cT-uT instead of cT.

Hence the corresponding frequency denoted by f1' its given by

f1'= c/ T(c-u)= cf/c-u

= f/ 1-(c/u)........ (1)

which represents and Apparent increase in the frequency if the source is moving away from the observer we put - u in place of u in equation( 1st) and get the corresponding f1'

in the second case When the source is stationary but the observer is moving with the velocity v from the source the wave is speed relative to the observer is not c but c-v and hence apparent frequency in the case is.

$f \tiny{2} = \frac{c - v}{ \lambda}$

= c-v/c/f

= c-v/c*f

$1 - \frac{v}{c} f \: .......(2)$

f.......(2)

now both observer and source are moving

$\bf{\lambda \tiny{1} = \frac{c}{f \tiny{1}} }$

= $\frac{c - u}{ \lambda}$