ARDMS SPI Exam | 2026-2027 Edition | 150 Questions & Correct Answers
ARDMS SPI EXAM — LATEST REAL EXAM 150
QUESTIONS AND CORRECT ANSWERS
A+ Grade | Brand New Version | 2026-2027 ARDMS Aligned
Sonography Principles and Instrumentation (SPI) — Comprehensive Physics and Instrumentation
Application
Examination structure: 150 multiple-choice questions across 8 sections, mirroring the 2026-2027 ARDMS Sonography
Principles and Instrumentation (SPI) content outline. Cognitive distribution: approximately 30% recall, 50% application,
and 20% analysis. Question styles include direct recall, scenario-based reasoning, and physics-formula calculations
(attenuation, wavelength, Doppler, resolution, RI/PI, Bernoulli). For each question, exactly one option is correct; rationales
explain why the correct answer is right and why the distractors reflect common SPI errors. Recommended time: 3 hours.
Passing benchmark: 75% correct.
Section 1: Ultrasound Physics Fundamentals (Q1–Q25)
Q1. A sonographer adjusts the imaging frequency from 5 MHz to 2.5 MHz on the same transducer.
Which of the following best describes the effect on wavelength in soft tissue (c = 1540 m/s)?
A. Wavelength is halved, from 0.31 mm to 0.154 mm
B. Wavelength doubles, from 0.308 mm to 0.616 mm [CORRECT]
C. Wavelength remains unchanged because frequency does not affect wavelength
D. Wavelength quadruples because wavelength is inversely proportional to the square of frequency
Correct Answer: B
Rationale: Wavelength (λ) = propagation speed / frequency. At 5 MHz in soft tissue, λ = 1540 m/s ÷ 5,000,000 s ¹ = 0.308
mm. At 2.5 MHz, λ = 1540 ÷ 2,500,000 = 0.616 mm. Wavelength therefore doubles when frequency is halved, because
wavelength is inversely proportional to frequency (assuming the medium is unchanged). Choices A, C, and D reflect the
most common SPI errors: halving instead of doubling, denying the inverse relationship, or inventing a squared relationship
that does not exist.
Q2. Which acoustic variable is most directly responsible for the propagation of a longitudinal
ultrasound wave through soft tissue?
A. Temperature
B. Particle velocity (motion of molecules parallel to beam direction) [CORRECT]
C. Electrical current within the medium
D. Magnetic flux density at the transducer face
Correct Answer: B
Rationale: The three acoustic variables are pressure, density, and particle motion. In a longitudinal (compressional) wave
such as diagnostic ultrasound, molecules oscillate parallel to the direction of propagation, transmitting energy by
compression and rarefaction. Temperature, electrical current, and magnetic flux are not acoustic variables and are not
responsible for sound transmission. Choice B correctly identifies particle motion as the propagating mechanism, while the
distractors represent physics domains unrelated to acoustic wave transmission.
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,ARDMS SPI Exam | 2026-2027 Edition | 150 Questions & Correct Answers
Q3. A pulsed-wave transducer emits a 4-cycle pulse at a frequency of 5 MHz. What is the spatial
pulse length (SPL) in soft tissue?
A. 0.308 mm
B. 0.77 mm
C. 1.232 mm [CORRECT]
D. 2.0 mm
Correct Answer: C
Rationale: SPL = number of cycles × wavelength. Wavelength at 5 MHz in soft tissue = 1540 m/s ÷ 5,000,000 s ¹ = 0.308
mm. SPL = 4 × 0.308 = 1.232 mm. Choice A is a single wavelength, choice B doubles it, and choice D is a fabricated value.
Only choice C applies the SPL formula correctly, a foundational calculation tested on the 2026-2027 SPI exam because SPL
directly determines axial resolution (axial resolution = SPL/2).
Q4. Which statement correctly compares the propagation speeds of ultrasound in air, soft tissue,
and bone?
A. Air > soft tissue > bone
B. Bone > air > soft tissue
C. Bone > soft tissue > air [CORRECT]
D. Soft tissue > bone > air
Correct Answer: C
Rationale: Typical propagation speeds are: air ≈ 330 m/s, soft tissue (average) ≈ 1540 m/s, and bone ≈ 4080 m/s. Speed
depends on the medium's stiffness (√(Bulk modulus / density)); stiffer media transmit sound faster. Therefore bone > soft
tissue > air, choice C. The distractors invert the order, a common SPI error caused by confusing density with stiffness;
although bone is denser than soft tissue, its far greater stiffness dominates the speed equation.
Q5. The intensity of an ultrasound beam is increased by a factor of 4. By what factor does the
amplitude of the wave increase?
A. 2 [CORRECT]
B. 4
C. 8
D. 16
Correct Answer: A
Rationale: Intensity is proportional to amplitude squared (I ∝ A²). Conversely, amplitude is proportional to the square root
of intensity (A ∝ √I). If intensity increases by 4×, amplitude increases by √4 = 2×. Choice A is correct. Choices B, C, and D
confuse the squared relationship, treating amplitude-intensity as linear, cubic, or quartic, which are common errors seen on
the SPI exam when candidates misapply the power-intensity-amplitude relationships.
Q6. An ultrasound pulse travels through 10 cm of soft tissue at 3 MHz. What is the total attenuation
(in dB) along this path? (Use the soft-tissue attenuation coefficient of 0.5 dB/cm/MHz.)
A. 7.5 dB
B. 15 dB [CORRECT]
C. 30 dB
D. 150 dB
Correct Answer: B
Rationale: Total attenuation = frequency (MHz) × distance (cm) × 0.5 dB/cm/MHz = 3 × 10 × 0.5 = 15 dB. Choice B is
correct. Choice A omits the 0.5 coefficient (3×10×2.5), choice C doubles the coefficient, and choice D forgets the 0.5 and
treats it as 1.0. The attenuation formula is one of the most frequently tested calculations on the SPI exam.
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,ARDMS SPI Exam | 2026-2027 Edition | 150 Questions & Correct Answers
Q7. Which phenomenon is primarily responsible for converting acoustic energy into heat within
soft tissue during diagnostic ultrasound?
A. Specular reflection at large smooth interfaces
B. Rayleigh scattering from red blood cells
C. Absorption by macromolecular relaxation processes [CORRECT]
D. Refraction at oblique tissue boundaries
Correct Answer: C
Rationale: Absorption is the dominant mechanism by which ultrasound energy is converted to heat. It occurs through
macromolecular relaxation, where tissue molecules absorb energy and dissipate it as thermal energy. Reflection, scattering,
and refraction redirect acoustic energy but do not directly convert it to heat. Choice C is correct. This principle underpins the
Thermal Index (TI) on the output display and is central to ALARA-based safety decisions in the 2026-2027 ARDMS content
outline.
Q8. A sound wave strikes a smooth interface between two media at exactly 90 degrees (normal
incidence). Which statement best describes the resulting reflection?
A. Diffuse reflection occurs in all directions regardless of impedance mismatch
B. Specular reflection occurs, with the reflected wave returning back along the incident path [CORRECT]
C. No reflection can occur at normal incidence regardless of impedance
D. Refraction will dominate over reflection at normal incidence
Correct Answer: B
Rationale: At normal incidence (0° from perpendicular) on a smooth, large interface, specular reflection occurs: the angle of
reflection equals the angle of incidence (both 0°), so the reflected wave returns along the incident path. The fraction reflected
depends on the impedance mismatch (RPC formula). Diffuse reflection requires a rough interface; refraction requires oblique
incidence with different propagation speeds. Choice B is correct.
Q9. Using Snell's Law, sound traveling from soft tissue (c1 = 1540 m/s) into a medium with c2 =
2000 m/s at an incident angle of 30° will produce a transmitted beam that:
A. Bends toward the normal because c2 > c1
B. Bends away from the normal because c2 > c1 [CORRECT]
C. Continues straight without bending because the angle is below the critical angle
D. Undergoes total internal reflection because c2 > c1
Correct Answer: B
Rationale: Snell's Law: sin(θᵢ)/c1 = sin(θ )/c2. When the transmitted medium has a higher speed (c2 > c1), sin(θ ) >
sin(θᵢ), so the transmitted angle is larger and the beam bends away from the normal. Choice B is correct. Total internal
reflection requires travel from a faster to a slower medium, ruling out choice D. Choice A reverses the bending rule, and
choice C incorrectly states that no refraction occurs below a critical angle.
Q10. Rayleigh scattering occurs when the reflector is much smaller than the wavelength of
insonating ultrasound. The intensity of Rayleigh-scattered ultrasound is proportional to:
A. Frequency
B. Frequency squared
C. Frequency to the fourth power [CORRECT]
D. Frequency to the sixth power
Correct Answer: C
Rationale: Rayleigh scattering intensity ∝ f⁴ (frequency to the fourth power). This is why red blood cells (much smaller than
the diagnostic wavelength) scatter far more strongly at higher frequencies, and it is the physical basis of the high sensitivity
required for Doppler detection of slow flow. Choice C is correct. The f⁴ relationship is one of the most frequently
misremembered formulas on the SPI exam, with candidates often choosing f or f².
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, ARDMS SPI Exam | 2026-2027 Edition | 150 Questions & Correct Answers
Q11. Which of the following correctly defines the period of an ultrasound wave?
A. The number of cycles per second, expressed in MHz
B. The distance traveled by one complete cycle, in mm
C. The time required to complete one cycle, typically in microseconds [CORRECT]
D. The rate at which energy is transferred, in W/cm²
Correct Answer: C
Rationale: Period is the time for one complete cycle to occur, typically expressed in microseconds (μs) for diagnostic
ultrasound. Choice A defines frequency, choice B defines wavelength, and choice D defines intensity. Choice C is correct.
Period and frequency are reciprocals: period = 1/frequency. At 5 MHz, the period = 1/5,000,000 = 0.2 μs. Period is
determined entirely by the source (transducer) and does not change as the wave propagates through tissue.
Q12. Which acoustic parameter remains unchanged as an ultrasound wave propagates from soft
tissue into bone?
A. Propagation speed
B. Wavelength
C. Frequency [CORRECT]
D. Impedance
Correct Answer: C
Rationale: Frequency is determined by the sound source (the transducer) and is unchanged as the wave crosses media.
Propagation speed increases in bone (~4080 m/s vs. 1540 m/s in soft tissue); wavelength = c/f, so wavelength also increases;
acoustic impedance (ρ·c) likewise increases. Only frequency is preserved across the interface. Choice C is correct. This is a
foundational SPI concept tested both as direct recall and within refraction/attenuation scenarios.
Q13. Acoustic impedance (Z) is the product of tissue density (ρ) and propagation speed (c). What
is the impedance of soft tissue (ρ = 1060 kg/m³, c = 1540 m/s)?
A. 1.63 × 10⁶ Rayl
B. 1.63 MRayl (1.63 × 10⁶ kg/m²s) [CORRECT]
C. 0.69 MRayl
D. 16.3 MRayl
Correct Answer: B
Rationale: Z = ρ × c = 1060 × 1540 = 1,632,400 kg/(m²·s) ≈ 1.63 × 10⁶ Rayl = 1.63 MRayl. Choice B is correct. Choice A is
numerically correct but lacks standard MRayl units commonly tested; choice C inverts the calculation; choice D adds an
erroneous order of magnitude. Impedance is central to reflection coefficient calculations (IRC = ((Z2-Z1)/(Z2+Z1))²) used
throughout SPI imaging physics.
Q14. What is the intensity reflection coefficient (IRC) at a soft tissue / air interface? (Z soft tissue ≈
1.63 MRayl, Z air ≈ 0.0004 MRayl.)
A. Approximately 0.1%
B. Approximately 11%
C. Approximately 50%
D. Greater than 99% [CORRECT]
Correct Answer: D
Rationale: IRC = ((Z2 − Z1)/(Z2 + Z1))². With Z1 = 1.63 and Z2 ≈ 0.0004 MRayl, the ratio (1.63 − 0.0004)/(1.63 + 0.0004)
≈ 0.9995, squared ≈ 0.999, so IRC ≈ 99.9%. This enormous impedance mismatch is why ultrasound cannot image through air
and why lung and bowel gas obscure visualization. Choice D is correct; the distractors dramatically underestimate the
reflection.
A+ Grade | Brand New Version Page 4
ARDMS SPI EXAM — LATEST REAL EXAM 150
QUESTIONS AND CORRECT ANSWERS
A+ Grade | Brand New Version | 2026-2027 ARDMS Aligned
Sonography Principles and Instrumentation (SPI) — Comprehensive Physics and Instrumentation
Application
Examination structure: 150 multiple-choice questions across 8 sections, mirroring the 2026-2027 ARDMS Sonography
Principles and Instrumentation (SPI) content outline. Cognitive distribution: approximately 30% recall, 50% application,
and 20% analysis. Question styles include direct recall, scenario-based reasoning, and physics-formula calculations
(attenuation, wavelength, Doppler, resolution, RI/PI, Bernoulli). For each question, exactly one option is correct; rationales
explain why the correct answer is right and why the distractors reflect common SPI errors. Recommended time: 3 hours.
Passing benchmark: 75% correct.
Section 1: Ultrasound Physics Fundamentals (Q1–Q25)
Q1. A sonographer adjusts the imaging frequency from 5 MHz to 2.5 MHz on the same transducer.
Which of the following best describes the effect on wavelength in soft tissue (c = 1540 m/s)?
A. Wavelength is halved, from 0.31 mm to 0.154 mm
B. Wavelength doubles, from 0.308 mm to 0.616 mm [CORRECT]
C. Wavelength remains unchanged because frequency does not affect wavelength
D. Wavelength quadruples because wavelength is inversely proportional to the square of frequency
Correct Answer: B
Rationale: Wavelength (λ) = propagation speed / frequency. At 5 MHz in soft tissue, λ = 1540 m/s ÷ 5,000,000 s ¹ = 0.308
mm. At 2.5 MHz, λ = 1540 ÷ 2,500,000 = 0.616 mm. Wavelength therefore doubles when frequency is halved, because
wavelength is inversely proportional to frequency (assuming the medium is unchanged). Choices A, C, and D reflect the
most common SPI errors: halving instead of doubling, denying the inverse relationship, or inventing a squared relationship
that does not exist.
Q2. Which acoustic variable is most directly responsible for the propagation of a longitudinal
ultrasound wave through soft tissue?
A. Temperature
B. Particle velocity (motion of molecules parallel to beam direction) [CORRECT]
C. Electrical current within the medium
D. Magnetic flux density at the transducer face
Correct Answer: B
Rationale: The three acoustic variables are pressure, density, and particle motion. In a longitudinal (compressional) wave
such as diagnostic ultrasound, molecules oscillate parallel to the direction of propagation, transmitting energy by
compression and rarefaction. Temperature, electrical current, and magnetic flux are not acoustic variables and are not
responsible for sound transmission. Choice B correctly identifies particle motion as the propagating mechanism, while the
distractors represent physics domains unrelated to acoustic wave transmission.
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,ARDMS SPI Exam | 2026-2027 Edition | 150 Questions & Correct Answers
Q3. A pulsed-wave transducer emits a 4-cycle pulse at a frequency of 5 MHz. What is the spatial
pulse length (SPL) in soft tissue?
A. 0.308 mm
B. 0.77 mm
C. 1.232 mm [CORRECT]
D. 2.0 mm
Correct Answer: C
Rationale: SPL = number of cycles × wavelength. Wavelength at 5 MHz in soft tissue = 1540 m/s ÷ 5,000,000 s ¹ = 0.308
mm. SPL = 4 × 0.308 = 1.232 mm. Choice A is a single wavelength, choice B doubles it, and choice D is a fabricated value.
Only choice C applies the SPL formula correctly, a foundational calculation tested on the 2026-2027 SPI exam because SPL
directly determines axial resolution (axial resolution = SPL/2).
Q4. Which statement correctly compares the propagation speeds of ultrasound in air, soft tissue,
and bone?
A. Air > soft tissue > bone
B. Bone > air > soft tissue
C. Bone > soft tissue > air [CORRECT]
D. Soft tissue > bone > air
Correct Answer: C
Rationale: Typical propagation speeds are: air ≈ 330 m/s, soft tissue (average) ≈ 1540 m/s, and bone ≈ 4080 m/s. Speed
depends on the medium's stiffness (√(Bulk modulus / density)); stiffer media transmit sound faster. Therefore bone > soft
tissue > air, choice C. The distractors invert the order, a common SPI error caused by confusing density with stiffness;
although bone is denser than soft tissue, its far greater stiffness dominates the speed equation.
Q5. The intensity of an ultrasound beam is increased by a factor of 4. By what factor does the
amplitude of the wave increase?
A. 2 [CORRECT]
B. 4
C. 8
D. 16
Correct Answer: A
Rationale: Intensity is proportional to amplitude squared (I ∝ A²). Conversely, amplitude is proportional to the square root
of intensity (A ∝ √I). If intensity increases by 4×, amplitude increases by √4 = 2×. Choice A is correct. Choices B, C, and D
confuse the squared relationship, treating amplitude-intensity as linear, cubic, or quartic, which are common errors seen on
the SPI exam when candidates misapply the power-intensity-amplitude relationships.
Q6. An ultrasound pulse travels through 10 cm of soft tissue at 3 MHz. What is the total attenuation
(in dB) along this path? (Use the soft-tissue attenuation coefficient of 0.5 dB/cm/MHz.)
A. 7.5 dB
B. 15 dB [CORRECT]
C. 30 dB
D. 150 dB
Correct Answer: B
Rationale: Total attenuation = frequency (MHz) × distance (cm) × 0.5 dB/cm/MHz = 3 × 10 × 0.5 = 15 dB. Choice B is
correct. Choice A omits the 0.5 coefficient (3×10×2.5), choice C doubles the coefficient, and choice D forgets the 0.5 and
treats it as 1.0. The attenuation formula is one of the most frequently tested calculations on the SPI exam.
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,ARDMS SPI Exam | 2026-2027 Edition | 150 Questions & Correct Answers
Q7. Which phenomenon is primarily responsible for converting acoustic energy into heat within
soft tissue during diagnostic ultrasound?
A. Specular reflection at large smooth interfaces
B. Rayleigh scattering from red blood cells
C. Absorption by macromolecular relaxation processes [CORRECT]
D. Refraction at oblique tissue boundaries
Correct Answer: C
Rationale: Absorption is the dominant mechanism by which ultrasound energy is converted to heat. It occurs through
macromolecular relaxation, where tissue molecules absorb energy and dissipate it as thermal energy. Reflection, scattering,
and refraction redirect acoustic energy but do not directly convert it to heat. Choice C is correct. This principle underpins the
Thermal Index (TI) on the output display and is central to ALARA-based safety decisions in the 2026-2027 ARDMS content
outline.
Q8. A sound wave strikes a smooth interface between two media at exactly 90 degrees (normal
incidence). Which statement best describes the resulting reflection?
A. Diffuse reflection occurs in all directions regardless of impedance mismatch
B. Specular reflection occurs, with the reflected wave returning back along the incident path [CORRECT]
C. No reflection can occur at normal incidence regardless of impedance
D. Refraction will dominate over reflection at normal incidence
Correct Answer: B
Rationale: At normal incidence (0° from perpendicular) on a smooth, large interface, specular reflection occurs: the angle of
reflection equals the angle of incidence (both 0°), so the reflected wave returns along the incident path. The fraction reflected
depends on the impedance mismatch (RPC formula). Diffuse reflection requires a rough interface; refraction requires oblique
incidence with different propagation speeds. Choice B is correct.
Q9. Using Snell's Law, sound traveling from soft tissue (c1 = 1540 m/s) into a medium with c2 =
2000 m/s at an incident angle of 30° will produce a transmitted beam that:
A. Bends toward the normal because c2 > c1
B. Bends away from the normal because c2 > c1 [CORRECT]
C. Continues straight without bending because the angle is below the critical angle
D. Undergoes total internal reflection because c2 > c1
Correct Answer: B
Rationale: Snell's Law: sin(θᵢ)/c1 = sin(θ )/c2. When the transmitted medium has a higher speed (c2 > c1), sin(θ ) >
sin(θᵢ), so the transmitted angle is larger and the beam bends away from the normal. Choice B is correct. Total internal
reflection requires travel from a faster to a slower medium, ruling out choice D. Choice A reverses the bending rule, and
choice C incorrectly states that no refraction occurs below a critical angle.
Q10. Rayleigh scattering occurs when the reflector is much smaller than the wavelength of
insonating ultrasound. The intensity of Rayleigh-scattered ultrasound is proportional to:
A. Frequency
B. Frequency squared
C. Frequency to the fourth power [CORRECT]
D. Frequency to the sixth power
Correct Answer: C
Rationale: Rayleigh scattering intensity ∝ f⁴ (frequency to the fourth power). This is why red blood cells (much smaller than
the diagnostic wavelength) scatter far more strongly at higher frequencies, and it is the physical basis of the high sensitivity
required for Doppler detection of slow flow. Choice C is correct. The f⁴ relationship is one of the most frequently
misremembered formulas on the SPI exam, with candidates often choosing f or f².
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, ARDMS SPI Exam | 2026-2027 Edition | 150 Questions & Correct Answers
Q11. Which of the following correctly defines the period of an ultrasound wave?
A. The number of cycles per second, expressed in MHz
B. The distance traveled by one complete cycle, in mm
C. The time required to complete one cycle, typically in microseconds [CORRECT]
D. The rate at which energy is transferred, in W/cm²
Correct Answer: C
Rationale: Period is the time for one complete cycle to occur, typically expressed in microseconds (μs) for diagnostic
ultrasound. Choice A defines frequency, choice B defines wavelength, and choice D defines intensity. Choice C is correct.
Period and frequency are reciprocals: period = 1/frequency. At 5 MHz, the period = 1/5,000,000 = 0.2 μs. Period is
determined entirely by the source (transducer) and does not change as the wave propagates through tissue.
Q12. Which acoustic parameter remains unchanged as an ultrasound wave propagates from soft
tissue into bone?
A. Propagation speed
B. Wavelength
C. Frequency [CORRECT]
D. Impedance
Correct Answer: C
Rationale: Frequency is determined by the sound source (the transducer) and is unchanged as the wave crosses media.
Propagation speed increases in bone (~4080 m/s vs. 1540 m/s in soft tissue); wavelength = c/f, so wavelength also increases;
acoustic impedance (ρ·c) likewise increases. Only frequency is preserved across the interface. Choice C is correct. This is a
foundational SPI concept tested both as direct recall and within refraction/attenuation scenarios.
Q13. Acoustic impedance (Z) is the product of tissue density (ρ) and propagation speed (c). What
is the impedance of soft tissue (ρ = 1060 kg/m³, c = 1540 m/s)?
A. 1.63 × 10⁶ Rayl
B. 1.63 MRayl (1.63 × 10⁶ kg/m²s) [CORRECT]
C. 0.69 MRayl
D. 16.3 MRayl
Correct Answer: B
Rationale: Z = ρ × c = 1060 × 1540 = 1,632,400 kg/(m²·s) ≈ 1.63 × 10⁶ Rayl = 1.63 MRayl. Choice B is correct. Choice A is
numerically correct but lacks standard MRayl units commonly tested; choice C inverts the calculation; choice D adds an
erroneous order of magnitude. Impedance is central to reflection coefficient calculations (IRC = ((Z2-Z1)/(Z2+Z1))²) used
throughout SPI imaging physics.
Q14. What is the intensity reflection coefficient (IRC) at a soft tissue / air interface? (Z soft tissue ≈
1.63 MRayl, Z air ≈ 0.0004 MRayl.)
A. Approximately 0.1%
B. Approximately 11%
C. Approximately 50%
D. Greater than 99% [CORRECT]
Correct Answer: D
Rationale: IRC = ((Z2 − Z1)/(Z2 + Z1))². With Z1 = 1.63 and Z2 ≈ 0.0004 MRayl, the ratio (1.63 − 0.0004)/(1.63 + 0.0004)
≈ 0.9995, squared ≈ 0.999, so IRC ≈ 99.9%. This enormous impedance mismatch is why ultrasound cannot image through air
and why lung and bowel gas obscure visualization. Choice D is correct; the distractors dramatically underestimate the
reflection.
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