ARDMS SPI EXAM VERSION A 2026 LATEST | Sonography
Principles & Instrumentation Actual Exam | Questions &
Verified Answers | Pass Guaranteed - A+ Graded
Section A1: Ultrasound Physics Fundamentals
Q1: A sound wave traveling through soft tissue has a frequency of 5 MHz. If the
propagation speed in soft tissue is 1540 m/s, what is the wavelength?
A. 0.154 mm
B. 0.308 mm [CORRECT]
C. 0.616 mm
D. 1.54 mm
Correct Answer: B
Rationale: On the ARDMS SPI exam (Version A), remember wavelength = propagation
speed / frequency. So 1540 m/s ÷ 5,000,000 Hz = 0.000308 m = 0.308 mm. That's the
fundamental relationship that underlies everything about resolution and penetration in
diagnostic ultrasound.
,Q2: The period of an ultrasound wave is defined as:
A. The distance between two consecutive compressions
B. The time it takes for one complete cycle to occur [CORRECT]
C. The maximum pressure variation from baseline
D. The rate at which energy passes through a unit area
Correct Answer: B
Rationale: That's right because of the physics principle that period is the time
component (seconds per cycle), while wavelength is the spatial component (distance
per cycle). A common SPI trap is confusing these two—period is time, wavelength is
distance, and they're inversely related to frequency.
Q3: As ultrasound frequency increases, which of the following occurs?
A. Wavelength increases and penetration increases
B. Wavelength decreases and axial resolution improves [CORRECT]
C. Propagation speed increases in soft tissue
D. Attenuation decreases and imaging depth increases
Correct Answer: B
Rationale: The correct adjustment is understanding that higher frequency means
shorter wavelength (λ = c/f), and shorter wavelength means better axial resolution since
,axial resolution equals spatial pulse length (approximately half the pulse length). But the
tradeoff is increased attenuation, so penetration suffers—that's the classic frequency
compromise tested on every SPI exam.
Q4: A pulse of ultrasound has a pulse duration of 2 microseconds and contains 3
cycles. What is the frequency of the transducer?
A. 1.0 MHz
B. 1.5 MHz [CORRECT]
C. 3.0 MHz
D. 6.0 MHz
Correct Answer: B
Rationale: On the ARDMS SPI exam (Version A), remember frequency = number of
cycles / pulse duration. So 3 cycles ÷ 2 microseconds (2 × 10⁻⁶ seconds) = 1.5 × 10⁶ Hz
= 1.5 MHz. This tests your understanding that pulse duration depends on both the
number of cycles and the frequency.
Q5: The attenuation coefficient in soft tissue is approximately:
A. 0.5 dB/cm/MHz [CORRECT]
B. 1.0 dB/cm/MHz
C. 2.0 dB/cm/MHz
, D. 3.5 dB/cm/MHz
Correct Answer: A
Rationale: That's right because of the physics principle that soft tissue attenuates
ultrasound at roughly 0.5 dB per centimeter per MHz. So a 5 MHz transducer loses
about 2.5 dB per cm of travel. This is the standard approximation used for imaging
depth calculations and TGC settings.
Q6: A 7 MHz transducer is imaging to a depth of 4 cm in soft tissue. Approximately how
much total attenuation occurs for the round trip?
A. 14 dB
B. 28 dB [CORRECT]
C. 56 dB
D. 112 dB
Correct Answer: B
Rationale: The correct calculation uses the round-trip distance (8 cm total) × attenuation
coefficient (0.5 dB/cm/MHz) × frequency (7 MHz) = 28 dB. A common SPI trap is
forgetting the round trip—sound travels to the reflector AND back, so you double the
one-way distance.
Principles & Instrumentation Actual Exam | Questions &
Verified Answers | Pass Guaranteed - A+ Graded
Section A1: Ultrasound Physics Fundamentals
Q1: A sound wave traveling through soft tissue has a frequency of 5 MHz. If the
propagation speed in soft tissue is 1540 m/s, what is the wavelength?
A. 0.154 mm
B. 0.308 mm [CORRECT]
C. 0.616 mm
D. 1.54 mm
Correct Answer: B
Rationale: On the ARDMS SPI exam (Version A), remember wavelength = propagation
speed / frequency. So 1540 m/s ÷ 5,000,000 Hz = 0.000308 m = 0.308 mm. That's the
fundamental relationship that underlies everything about resolution and penetration in
diagnostic ultrasound.
,Q2: The period of an ultrasound wave is defined as:
A. The distance between two consecutive compressions
B. The time it takes for one complete cycle to occur [CORRECT]
C. The maximum pressure variation from baseline
D. The rate at which energy passes through a unit area
Correct Answer: B
Rationale: That's right because of the physics principle that period is the time
component (seconds per cycle), while wavelength is the spatial component (distance
per cycle). A common SPI trap is confusing these two—period is time, wavelength is
distance, and they're inversely related to frequency.
Q3: As ultrasound frequency increases, which of the following occurs?
A. Wavelength increases and penetration increases
B. Wavelength decreases and axial resolution improves [CORRECT]
C. Propagation speed increases in soft tissue
D. Attenuation decreases and imaging depth increases
Correct Answer: B
Rationale: The correct adjustment is understanding that higher frequency means
shorter wavelength (λ = c/f), and shorter wavelength means better axial resolution since
,axial resolution equals spatial pulse length (approximately half the pulse length). But the
tradeoff is increased attenuation, so penetration suffers—that's the classic frequency
compromise tested on every SPI exam.
Q4: A pulse of ultrasound has a pulse duration of 2 microseconds and contains 3
cycles. What is the frequency of the transducer?
A. 1.0 MHz
B. 1.5 MHz [CORRECT]
C. 3.0 MHz
D. 6.0 MHz
Correct Answer: B
Rationale: On the ARDMS SPI exam (Version A), remember frequency = number of
cycles / pulse duration. So 3 cycles ÷ 2 microseconds (2 × 10⁻⁶ seconds) = 1.5 × 10⁶ Hz
= 1.5 MHz. This tests your understanding that pulse duration depends on both the
number of cycles and the frequency.
Q5: The attenuation coefficient in soft tissue is approximately:
A. 0.5 dB/cm/MHz [CORRECT]
B. 1.0 dB/cm/MHz
C. 2.0 dB/cm/MHz
, D. 3.5 dB/cm/MHz
Correct Answer: A
Rationale: That's right because of the physics principle that soft tissue attenuates
ultrasound at roughly 0.5 dB per centimeter per MHz. So a 5 MHz transducer loses
about 2.5 dB per cm of travel. This is the standard approximation used for imaging
depth calculations and TGC settings.
Q6: A 7 MHz transducer is imaging to a depth of 4 cm in soft tissue. Approximately how
much total attenuation occurs for the round trip?
A. 14 dB
B. 28 dB [CORRECT]
C. 56 dB
D. 112 dB
Correct Answer: B
Rationale: The correct calculation uses the round-trip distance (8 cm total) × attenuation
coefficient (0.5 dB/cm/MHz) × frequency (7 MHz) = 28 dB. A common SPI trap is
forgetting the round trip—sound travels to the reflector AND back, so you double the
one-way distance.
