Echocardiography Board Review: 500 Multiple Choice Questions With Discussion

Chapter 3

Questions

1.  41. As transducer frequency increases, backscatter strength:

1.  A. Decreases

2.  B. Increases

3.  C. Does not change

4.  D. Refracts

2.  42. If an echo arrives 39 µs after a pulse has been emitted, at what depth should the reflecting object be on the scan line?

1.  A. 3 cm

2.  B. 6 cm

3.  C. 1 cm

4.  D. None of the above

3.  43. The Doppler shift produced by an object moving at a speed of 1 m/s toward the transducer emitting ultrasound at 2 MHz would be:

1.  A. 2.6 kHz

2.  B. 1.3 kHz

3.  C. 1 MHz

4.  D. 200 Hz

4.  44. In the above example, the reflected ultrasound will have a frequency of:

1.  A. 2 002 600 Hz

2.  B. 1 998 700 Hz

3.  C. 1 000 000 Hz

4.  D. 2 MHz

5.  45. Reflected ultrasound from an object moving away from the sound source will have a frequency:

1.  A. Higher than original sound

2.  B. Lower than the original sound

3.  C. Same as the original sound

4.  D. Variable, depending on source of sound and velocity of the moving object

6.  46. Reflected ultrasound from an object moving perpendicular to the sound source will have a frequency:

1.  A. Higher than original sound

2.  B. Lower than the original sound

3.  C. Same as the original sound

4.  D. Variable, depending on source of sound and velocity of the moving object

7.  47. Doppler shift frequency is independent of:

1.  A. Operating frequency

2.  B. Doppler angle

3.  C. Propagation speed

4.  D. Amplitude

8.  48. On a continuous wave Doppler display, amplitude is represented by:

1.  A. Brightness of the signal

2.  B. Vertical extent of the signal

3.  C. Width of the signal

4.  D. None of the above

9.  49. Doppler signals from the myocardium, compared with those from the blood pool, display:

1.  A. Lower velocity

2.  B. Greater amplitude

3.  C. Both of the above

4.  D. None of the above

10. 50. Doing which of the following modifications to the Doppler processing will allow myocardial velocities to be recorded selectively compared with blood pool velocities?

1.  A. A band pass filter that allows low velocities

2.  B. A band pass filter that allows high amplitude signals

3.  C. Both

4.  D. Neither

11. 51. If the propagation speed is 1.6 mm/µs and the pulse round trip time is 5 µs, the distance to the reflector is:

1.  A. 8 mm

2.  B. 4 mm

3.  C. 10 mm

4.  D. Cannot be determined

12. 52. How long after a pulse is sent out by a transducer does an echo from an object at a depth of 5 cm return?

1.  A. 13 µs

2.  B. 65 µs

3.  C. 5 µs

4.  D. Cannot be determined

13. 53. For soft tissues, the attenuation coefficient at 3 MHz is:

1.  A. 1 dB/cm

2.  B. 6 dB/cm

3.  C. 1.5 dB/cm

4.  D. 3 dB/cm

14. 54. If the density of a medium is 1000 kg/m3 and the propagation speed is 1540 m/s, the impedance is:

1.  A. 1 540 000 rayls

2.  B. 770 000 rayls

3.  C. 3 080 000 rayls

4.  D. Cannot be determined

15. 55. If the propagation speed through medium 2 is greater than the propagation speed through medium 1 the transmission angle will be _______ the incidence angle.

1.  A. Lesser

2.  B. Greater

3.  C. Equal to

4.  D. Cannot be determined

16. 56. If amplitude is doubled, intensity is:

1.  A. Halved

2.  B. Quadrupled

3.  C. Remains the same

4.  D. Tripled

17. 57. If both power and area are doubled, intensity is:

1.  A. Doubled

2.  B. Unchanged

3.  C. Halved

4.  D. Tripled

18. 58. Flow resistance in a vessel depends on:

1.  A. Vessel length

2.  B. Vessel radius

3.  C. Blood viscosity

4.  D. All of the above

5.  E. None of the above

19. 59. Flow resistance decreases with an increase in:

1.  A. Vessel length

2.  B. Vessel radius

3.  C. Blood viscosity

4.  D. None of the above

20. 60. Flow resistance depends most strongly on:

1.  A. Vessel length

2.  B. Vessel radius

3.  C. Blood viscosity

4.  D. All of the above

Answers for chapter 3

1.  41. Answer: B.

Higher frequency is associated with shorter wavelengths. Shorter wavelengths are more readily reflected compared to longer wavelengths.

2.  42. Answer: A.

Ultrasound takes 6.5 ms to travel 1 cm in the tissues assuming a transmission speed of 1540 m/s. Travel time for 6 cm is 39 µs; hence the object is 3 cm deep.

3.  43. Answer: A.

Fd = (2FoVcos of incident angle)/C where Fd is the Doppler shift, V is the velocity and C is the speed of sound in the medium. In this example, Fd = (2 × 2 000 000 × 1 × 1)/ 1540 = 2600 Hz or 2.6 kHz. For each MHz of emitted sound, a target velocity of 1 m/s will produce a Doppler shift of 1.3 kHz. Angle theta or incident angle is zero; hence cosine of that angle is 1.

4.  44. Answer: A.

As the object is moving directly toward the source of sound, the reflected sound will have a higher frequency and will equal Fo plus Fd.

5.  45. Answer: B.

Object moving away will produce a negative Doppler shift. Hence frequency of reflected ultrasound will be lower than transmitted frequency.

6.  46. Answer: C.

As the cosine of the incident angle of 90° is zero, the Doppler shift is zero (please look at Doppler equation in question 43). Because of the this angle dependence of the Doppler shift, the angle between the direction of motion of the object and the ultrasound beam has to be as close to zero as possible to record the true Doppler shift and hence the true velocity. Cosine of 0°is 1, cosine of 20° is 0.94, and cosine of 90° is 0. Angle correction is generally not used for intracardiac flows because of the three-dimensional nature of intracardiac flows and fallacies of assumed angles in contrast to flow in tubular structures such as blood vessels.

7.  47. Answer: D.

Please look up the Doppler equation in question 43. Note that Fd depends on ultrasound frequency, velocity of motion, direction of motion, and speed of sound in the medium but not amplitude or gain.

8.  48. Answer: A.

Amplitude is the strength of the returning signal. Vertical extent is the velocity of the object, and horizontal axis is the time axis and gives distribution or timing of the signal in the cardiac cycle.

9.  49. Answer: C.

Myocardium produces stronger or higher amplitude signals that have lower velocities compared to the blood pool.

10. 50. Answer: C.

The blood pool signals have higher velocity and lower amplitude compared with myocardial signals. Thus, filtering of higher velocity/lower amplitude signals will allow only low velocity/higher amplitude signals that come from the myocardium.

11. 51. Answer: B.

The distance to the reflector is calculated by the range equation. The formula is ½ (propagation speed (mm/µs) × round trip time (µs)). So solving the equation gives ½ (1.6 × 5) = 4 mm. In other words, in 5 µs , sound would have traveled 8 mm (to and fro distance). Hence depth is 4 mm.

12. 52. Answer: B.

The round trip travel time for 1 cm is 13 µs. Hence for an object at 5 cm, the travel time is 13 µs × 5 = 65 µs.

13. 53. Answer: C.

Attenuation coefficient in soft tissue is equivalent to ½ × frequency (MHz). In the above question ½ × 3 = 1.5 dB/cm. Multiplying this by the path length (cm) yields the attenuation (dB).

14. 54. Answer: A.

Impedance describes the relationship between acoustic pressure and the speed of particle vibrations in a sound wave. It is equal to the density of a medium × propagation speed. Solving the equation gives 1000 × 1540 = 1 540 000 rayls. Impedance is increased if the density of the medium is increased or the propagation speed is increased. Note that all units are in MKS (meter, kilogram, seconds). Hence unit of Rayls is (kg/m3) × (m/s) = kg/m2s.

15. 55. Answer: B.

When the propagation speed in medium 2 is greater than medium 1, the transmission angle will be greater than the incidence angle.

16. 56. Answer: B.

Intensity is the rate at which energy passes through a unit area. Intensity is equal to amplitude squared. Hence, if amplitude is doubled, intensity is quadrupled.

17. 57. Answer: B.

Intensity is given by the equation power (mW)/area (cm2). Hence if both power and area are doubled, intensity will remain the same as both numerator and denominator are multiplied by the same number.

18. 58. Answer: D.

Flow resistance is = (8 × length × viscosity)/(π × radius4). Hence flow resistance is directly proportional to length and viscosity and inversely proportional to the 4th power of the radius.

19. 59. Answer: B.

Flow resistance decreases with an increase in the vessel radius. Please refer to question 58 for the relationship. Resistance to flow and hence flow rate for a given driving pressure depends upon radius, length, and viscosity.

20. 60. Answer: B.

Flow resistance is inversely related to the 4th power of radius. Hence it is most strongly related to the vessel radius. Rα 1/r4.