Ultrasound, which consists of sound waves with frequencies above the human audible range, can be used to produce an image of the interior of a human body. Moreover, ultrasound can be used to measure the speed of the blood in the body; it does so by comparing the frequency of the ultrasound sent into the body with the frequency of the ultrasound reflected back to the body’s surface by the blood. As the blood pulses, this detected frequency varies.

Suppose that an ultrasound image of the arm of a patient shows an artery that is angled atθ=20°to the ultrasound’s line of travel (Fig. 17-47). Suppose also that the frequency of the ultrasound reflected by the blood in the artery is increased by a maximum of5495 Hzfrom the original ultrasound frequency of5.000000 MHz

(a) In Fig. 17-47, is the direction of the blood flow rightward or leftward?

(b) The speed of sound in the human arm is1540 m/s. What is the maximum speed of the blood?

(c) If angleuwere greater, would the reflected frequency be greater or less?

Short Answer

Expert verified
  1. The blood flow is in rightward direction.
  2. The maximum speed of the blood is0.90 m/s
  3. The reflected frequency is less for angle u being greater.

Step by step solution

01

The given data

  1. Speed of sound in the human arm,v=1540 m/s.
  2. Frequency of original ultrasound,f=5 MHzor5×106 Hz.
  3. Increased frequencyΔf=5495 Hz.
  4. Angle between human arm and ultrasound’s travel line, θ=200.
02

Understanding the concept of the Doppler Effect

Using the formula of the Doppler Effect, we can find the direction of the blood flow, and also, we can find the maximum speed of the blood.

Formula:

The frequency received by the observer according to Doppler’s Effect,

f+Δf=f(v+vxvvx) …(i)

03

a) Calculation of the direction of the blood flow

Since we are given that the frequency increases in Doppler shift, the blood is moving towards the right.

Therefore, the blood flow is rightward.

04

b) Calculation of maximum speed 

For the reception of the ultrasound by the blood and the subsequent remitting of the signal by the blood, we write the frequency relation as equation (ii):

f+Δff=ffv+vxvvx

Let,K=f+Δff

K=v+vxvvx(vvx)K=v+vxvKvxK=v+vxvxK+vx=vKvvx(K+1)=v(K1)

The velocity of the blood is expressed as,

vbloodcosθ0(K+1)=v(K1)vblood=v(K1)cosθ0(K+1)

Where, the value of K is found to be

K=f+Δff=(5×106 Hz)+(5495 Hz)(5×106 Hz)=1.001099

Hence, substituting the given values in the above equation , we get

vblood=(1540 m/s)(1.0010991)cos200(1.001099+1)vblood=0.90 m/s

Therefore, the maximum speed of the blood is 0.90 m/s.

05

c) Checking the behaviour of reflected frequency with respect to the angle

As the angle increases, the horizontal component of velocity decreases. Therefore, it is expected that the frequencyfdecreases whentheangle increases from0° to 90°.

Hence, the reflected frequency is less for larger angles.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sperm whale (Fig. 17-44a) vocalizes by producing a series of clicks. Actually, the whale makes only a single sound near the front of its head to start the series. Part of that sound then emerges from the head into the water to become the first click of the series. The rest of the sound travels backward through the spermaceti sac (a body of fat), reflects from the frontal sac (an air layer), and then travels forward through the spermaceti sac. When it reaches the distal sac (another air layer) at the front of the head, some of the sound escapes into the water to form the second click, and the rest is sent back through the spermaceti sac (and ends up forming later clicks).

Figure 17-44bshows a strip-chart recording of a series of clicks. A unit time interval of1.0 msis indicated on the chart. Assuming that the speed of sound in the spermaceti sac is1372 m/s, find the length of the spermaceti sac. From such a calculation, marine scientists estimate the length of a whale from its click series.

A certain loudspeaker system emits sound isotropically with a frequency of 2000 Hz and an intensity ofrole="math" localid="1661500478873" 0.960 mW/m2 at a distance ofrole="math" localid="1661501289787" 6.10 m . Assume that there are no reflections. (a) What is the intensity at 30.0 m? At 6.10 m, what are (b) the displacement amplitude and (c) the pressure amplitude?

Suppose that the sound level of a conversation in initially at an angry70dB and then drops to a soothing50dB. Assuming that the frequency of the sound is500Hz, determine the (a)Initial (b) Final sound intensities and(c)Initial and(d)Final sound wave amplitude

The sixth harmonic is set up in a pipe. (a) How many open ends does the pipe have (it has at least one)? (b) Is there a node, antinode, or some intermediate state at the midpoint?

In pipe A, the ratio of a particular harmonic frequency to the next lower harmonic frequency is 1.2.In pipeB, the ratio of a particular harmonic frequency to the next lower harmonic frequency is1.4. How many open ends are in (a) pipeAand (b) pipeB?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free