ARDMS SONOGRAPHY PRINCIPLES &
INSTRUMENTATION (SPI) EXAM Actual Exam
Complete Questions and Answers Detailed Rationales
Pass Guaranteed - A+ Graded
Total Questions: 50 | Time: 90 min | Pass: 80%
TABLE OF CONTENTS
Section 1 | Ultrasound Physics & Acoustic Wave Propagation | Q1 – Q10
Section 2 | Transducers & Beam Formation | Q11 – Q20
Section 3 | Image Formation, Resolution & Artifacts | Q21 – Q30
Section 4 | Doppler Principles & Hemodynamics | Q31 – Q40
Section 5 | Quality Assurance, Safety & Bioeffects | Q41 – Q50
Instructions: Choose the single best answer. Pass: 80% in 90 minutes.
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SECTION 1: ULTRASOUND PHYSICS & ACOUSTIC WAVE PROPAGATION Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A sonographer is imaging the liver and notices the system is set to 7 MHz. The speed of
sound in soft tissue is approximately 1540 m/s. What is the wavelength?
A. 0.11 mm
B. 0.15 mm
C. 0.22 mm ✓ CORRECT
D. 0.30 mm
Correct Answer: C
Rationale: Wavelength is calculated by dividing the speed of sound by the frequency, so
1540 m/s divided by 7,000,000 Hz equals 0.00022 m or 0.22 mm. This relationship is
fundamental because axial resolution is directly related to wavelength. A common
,mistake is dividing by 3.5 MHz, which yields 0.44 mm and confuses the calculation with
a different frequency.
Question 2 of 50
During an abdominal scan, the sonographer increases the output power from 20 dB to
50 dB. By what factor has the acoustic intensity increased?
A. 3 times
B. 10 times
C. 100 times
D. 1000 times ✓ CORRECT
Correct Answer: D
Rationale: A 30 dB increase in power corresponds to a 1000-fold increase in intensity
because decibels are logarithmic and every 10 dB represents a tenfold intensity change.
Intensity ratios use the formula 10^(dB difference/10). A tempting error is assuming a
linear relationship and selecting 30 times, but decibels do not scale linearly with
intensity.
Question 3 of 50
A pulse of ultrasound passes from fat (acoustic impedance 1.4) into muscle (acoustic
impedance 1.7). What percentage of the beam is reflected at the interface?
A. 0.01%
B. 1% ✓ CORRECT
C. 10%
D. 50%
Correct Answer: B
Rationale: The intensity reflection coefficient is calculated as [(Z2 − Z1)/(Z2 + Z1)]²,
which equals [(0.3)/(3.1)]² or approximately 0.0094, meaning about 1% of the beam is
reflected. Fat and muscle have relatively similar impedances, so only a small fraction
, reflects. Selecting 10% would require a much larger impedance mismatch, such as soft
tissue to bone or soft tissue to air.
Question 4 of 50
A sonographer is scanning a patient with ascites and notices the posterior liver appears
brighter than expected. This is caused by:
A. Decreased attenuation through the fluid causing acoustic enhancement ✓ CORRECT
B. Increased attenuation through the fluid
C. Reflection from the diaphragm
D. Refraction at the liver capsule
Correct Answer: A
Rationale: Fluid attenuates sound less than solid tissue, so more acoustic energy
reaches the posterior liver and returns to the transducer, creating acoustic
enhancement. Increased attenuation would darken rather than brighten the posterior
liver. Reflection from the diaphragm creates a mirror image artifact, and refraction at the
capsule does not cause generalized posterior brightness.
Question 5 of 50
The sonographer decreases the imaging depth from 12 cm to 6 cm. How does this
affect the pulse repetition frequency?
A. PRF must decrease to allow deeper listening
B. PRF can increase because the listening time is shorter ✓ CORRECT
C. PRF remains unchanged because it is set by the transducer
D. PRF decreases proportionally to the square of the depth
Correct Answer: C
Rationale: PRF is limited by the round-trip time required for the pulse to travel to the
maximum imaging depth and return, so halving the depth allows the system to double
the PRF. The transducer frequency does not dictate PRF, and the relationship is linear
INSTRUMENTATION (SPI) EXAM Actual Exam
Complete Questions and Answers Detailed Rationales
Pass Guaranteed - A+ Graded
Total Questions: 50 | Time: 90 min | Pass: 80%
TABLE OF CONTENTS
Section 1 | Ultrasound Physics & Acoustic Wave Propagation | Q1 – Q10
Section 2 | Transducers & Beam Formation | Q11 – Q20
Section 3 | Image Formation, Resolution & Artifacts | Q21 – Q30
Section 4 | Doppler Principles & Hemodynamics | Q31 – Q40
Section 5 | Quality Assurance, Safety & Bioeffects | Q41 – Q50
Instructions: Choose the single best answer. Pass: 80% in 90 minutes.
══════════════════════════════════════
SECTION 1: ULTRASOUND PHYSICS & ACOUSTIC WAVE PROPAGATION Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A sonographer is imaging the liver and notices the system is set to 7 MHz. The speed of
sound in soft tissue is approximately 1540 m/s. What is the wavelength?
A. 0.11 mm
B. 0.15 mm
C. 0.22 mm ✓ CORRECT
D. 0.30 mm
Correct Answer: C
Rationale: Wavelength is calculated by dividing the speed of sound by the frequency, so
1540 m/s divided by 7,000,000 Hz equals 0.00022 m or 0.22 mm. This relationship is
fundamental because axial resolution is directly related to wavelength. A common
,mistake is dividing by 3.5 MHz, which yields 0.44 mm and confuses the calculation with
a different frequency.
Question 2 of 50
During an abdominal scan, the sonographer increases the output power from 20 dB to
50 dB. By what factor has the acoustic intensity increased?
A. 3 times
B. 10 times
C. 100 times
D. 1000 times ✓ CORRECT
Correct Answer: D
Rationale: A 30 dB increase in power corresponds to a 1000-fold increase in intensity
because decibels are logarithmic and every 10 dB represents a tenfold intensity change.
Intensity ratios use the formula 10^(dB difference/10). A tempting error is assuming a
linear relationship and selecting 30 times, but decibels do not scale linearly with
intensity.
Question 3 of 50
A pulse of ultrasound passes from fat (acoustic impedance 1.4) into muscle (acoustic
impedance 1.7). What percentage of the beam is reflected at the interface?
A. 0.01%
B. 1% ✓ CORRECT
C. 10%
D. 50%
Correct Answer: B
Rationale: The intensity reflection coefficient is calculated as [(Z2 − Z1)/(Z2 + Z1)]²,
which equals [(0.3)/(3.1)]² or approximately 0.0094, meaning about 1% of the beam is
reflected. Fat and muscle have relatively similar impedances, so only a small fraction
, reflects. Selecting 10% would require a much larger impedance mismatch, such as soft
tissue to bone or soft tissue to air.
Question 4 of 50
A sonographer is scanning a patient with ascites and notices the posterior liver appears
brighter than expected. This is caused by:
A. Decreased attenuation through the fluid causing acoustic enhancement ✓ CORRECT
B. Increased attenuation through the fluid
C. Reflection from the diaphragm
D. Refraction at the liver capsule
Correct Answer: A
Rationale: Fluid attenuates sound less than solid tissue, so more acoustic energy
reaches the posterior liver and returns to the transducer, creating acoustic
enhancement. Increased attenuation would darken rather than brighten the posterior
liver. Reflection from the diaphragm creates a mirror image artifact, and refraction at the
capsule does not cause generalized posterior brightness.
Question 5 of 50
The sonographer decreases the imaging depth from 12 cm to 6 cm. How does this
affect the pulse repetition frequency?
A. PRF must decrease to allow deeper listening
B. PRF can increase because the listening time is shorter ✓ CORRECT
C. PRF remains unchanged because it is set by the transducer
D. PRF decreases proportionally to the square of the depth
Correct Answer: C
Rationale: PRF is limited by the round-trip time required for the pulse to travel to the
maximum imaging depth and return, so halving the depth allows the system to double
the PRF. The transducer frequency does not dictate PRF, and the relationship is linear
