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ARDMS SPI EXAM SONOGRAPHY PRINCIPLES 2026/2027
| Latest Real Exam 150 Questions | Brand New Version | A
Grade | Pass Guaranteed
Section 1: Sound Wave Physics & Propagation (Q1-25)
Question 1 A sonographer switches from a 3 MHz transducer to a 6 MHz transducer
to image a superficial structure. Assuming the propagation speed in soft tissue
remains constant, what happens to the wavelength?
A. It doubles
B. It is halved
C. It remains unchanged
D. It quadruples
Rationale: B. It is halved [CORRECT]. The relationship c = fλ dictates that if
propagation speed (c) is constant and frequency (f) doubles, wavelength (λ) must be
halved. Option A is incorrect because wavelength decreases with increasing
frequency. Option C is incorrect because wavelength is inversely proportional to
frequency. Option D is incorrect because the relationship is linear inverse, not
quadratic.
Correct Answer: B
Question 2 The average propagation speed of sound in soft tissue is approximately:
A. 330 m/s
B. 1540 m/s
C. 4080 m/s
D. 7700 m/s
Rationale: B. 1540 m/s [CORRECT]. This is the standard average value used in
ultrasound systems for soft tissue. Option A (330 m/s) is the speed in air. Option C
,2
(4080 m/s) is approximately the speed in bone. Option D (7700 m/s) is not a
physiologically relevant speed.
Correct Answer: B
Question 3 Two media have acoustic impedances of 1.6 MRayl and 1.7 MRayl. At
normal incidence, the percentage of reflected intensity is approximately:
A. 0.1%
B. 1%
C. 10%
D. 50%
Rationale: A. 0.1% [CORRECT]. The intensity reflection coefficient = [(Z2 - Z1)/(Z2 +
Z1)]² = [(0.1)/(3.3)]² ≈ 0.0009 or 0.09%. Option B overestimates by tenfold. Option C
and D are gross overestimates that would occur with much larger impedance
mismatches.
Correct Answer: A
Question 4 The half-value layer of a particular tissue is 2 cm. After the sound beam
travels 6 cm through this tissue, the intensity is reduced to what fraction of the
original?
A. 1/2
B. 1/4
C. 1/8
D. 1/16
Rationale: C. 1/8 [CORRECT]. Six centimeters represents three half-value layers (6 ÷
2 = 3). Intensity is reduced by (1/2)³ = 1/8. Option A represents one HVL. Option B
represents two HVLs. Option D represents four HVLs.
Correct Answer: C
,3
Question 5 A 2026 ARDMS SPI blueprint update emphasizes harmonized content
with CCI regarding:
A. Only obstetric ultrasound physics
B. Cross-disciplinary sonography physics standards applicable to all specialty areas
C. Only cardiac ultrasound instrumentation
D. Only vascular ultrasound Doppler
Rationale: B. Cross-disciplinary sonography physics standards applicable to all
specialty areas [CORRECT]. The 2026 update harmonized core physics content across
credentialing organizations. Options A, C, and D are incorrect because the
harmonization is comprehensive, not limited to a single specialty.
Correct Answer: B
Question 6 The period of a 7 MHz sound wave is:
A. 0.14 microseconds
B. 0.14 milliseconds
C. 1.43 microseconds
D. 1.43 milliseconds
Rationale: A. 0.14 microseconds [CORRECT]. Period (T) = 1/f = 1/7,000,000 s ≈ 0.143
× 10⁻⁶ s = 0.143 microseconds. Option C is incorrect because it is ten times larger
(1.43 microseconds). Options B and D use milliseconds, which are orders of
magnitude too large.
Correct Answer: A
Question 7 Which tissue has the HIGHEST attenuation coefficient at a given
ultrasound frequency?
, 4
A. Blood
B. Liver
C. Bone
D. Fat
Rationale: C. Bone [CORRECT]. Bone has the highest attenuation coefficient among
these tissues due to its density and stiffness. Blood (A) has very low attenuation. Liver
(B) and fat (D) have moderate attenuation, but bone attenuates sound significantly
more.
Correct Answer: C
Question 8 Acoustic impedance is calculated as:
A. Density × propagation speed
B. Density ÷ propagation speed
C. Frequency × wavelength
D. Pressure ÷ volume
Rationale: A. Density × propagation speed [CORRECT]. Acoustic impedance (Z) = ρ
× c, where ρ is density and c is propagation speed. Option B inverts the relationship.
Option C defines speed, not impedance. Option D is unrelated to acoustic
impedance.
Correct Answer: A
Question 9 At an interface between soft tissue (Z ≈ 1.5 MRayl) and air (Z ≈ 0.0004
MRayl), the percentage of reflected intensity is approximately:
A. 0.1%
B. 1%
C. 50%
D. 99.9%
ARDMS SPI EXAM SONOGRAPHY PRINCIPLES 2026/2027
| Latest Real Exam 150 Questions | Brand New Version | A
Grade | Pass Guaranteed
Section 1: Sound Wave Physics & Propagation (Q1-25)
Question 1 A sonographer switches from a 3 MHz transducer to a 6 MHz transducer
to image a superficial structure. Assuming the propagation speed in soft tissue
remains constant, what happens to the wavelength?
A. It doubles
B. It is halved
C. It remains unchanged
D. It quadruples
Rationale: B. It is halved [CORRECT]. The relationship c = fλ dictates that if
propagation speed (c) is constant and frequency (f) doubles, wavelength (λ) must be
halved. Option A is incorrect because wavelength decreases with increasing
frequency. Option C is incorrect because wavelength is inversely proportional to
frequency. Option D is incorrect because the relationship is linear inverse, not
quadratic.
Correct Answer: B
Question 2 The average propagation speed of sound in soft tissue is approximately:
A. 330 m/s
B. 1540 m/s
C. 4080 m/s
D. 7700 m/s
Rationale: B. 1540 m/s [CORRECT]. This is the standard average value used in
ultrasound systems for soft tissue. Option A (330 m/s) is the speed in air. Option C
,2
(4080 m/s) is approximately the speed in bone. Option D (7700 m/s) is not a
physiologically relevant speed.
Correct Answer: B
Question 3 Two media have acoustic impedances of 1.6 MRayl and 1.7 MRayl. At
normal incidence, the percentage of reflected intensity is approximately:
A. 0.1%
B. 1%
C. 10%
D. 50%
Rationale: A. 0.1% [CORRECT]. The intensity reflection coefficient = [(Z2 - Z1)/(Z2 +
Z1)]² = [(0.1)/(3.3)]² ≈ 0.0009 or 0.09%. Option B overestimates by tenfold. Option C
and D are gross overestimates that would occur with much larger impedance
mismatches.
Correct Answer: A
Question 4 The half-value layer of a particular tissue is 2 cm. After the sound beam
travels 6 cm through this tissue, the intensity is reduced to what fraction of the
original?
A. 1/2
B. 1/4
C. 1/8
D. 1/16
Rationale: C. 1/8 [CORRECT]. Six centimeters represents three half-value layers (6 ÷
2 = 3). Intensity is reduced by (1/2)³ = 1/8. Option A represents one HVL. Option B
represents two HVLs. Option D represents four HVLs.
Correct Answer: C
,3
Question 5 A 2026 ARDMS SPI blueprint update emphasizes harmonized content
with CCI regarding:
A. Only obstetric ultrasound physics
B. Cross-disciplinary sonography physics standards applicable to all specialty areas
C. Only cardiac ultrasound instrumentation
D. Only vascular ultrasound Doppler
Rationale: B. Cross-disciplinary sonography physics standards applicable to all
specialty areas [CORRECT]. The 2026 update harmonized core physics content across
credentialing organizations. Options A, C, and D are incorrect because the
harmonization is comprehensive, not limited to a single specialty.
Correct Answer: B
Question 6 The period of a 7 MHz sound wave is:
A. 0.14 microseconds
B. 0.14 milliseconds
C. 1.43 microseconds
D. 1.43 milliseconds
Rationale: A. 0.14 microseconds [CORRECT]. Period (T) = 1/f = 1/7,000,000 s ≈ 0.143
× 10⁻⁶ s = 0.143 microseconds. Option C is incorrect because it is ten times larger
(1.43 microseconds). Options B and D use milliseconds, which are orders of
magnitude too large.
Correct Answer: A
Question 7 Which tissue has the HIGHEST attenuation coefficient at a given
ultrasound frequency?
, 4
A. Blood
B. Liver
C. Bone
D. Fat
Rationale: C. Bone [CORRECT]. Bone has the highest attenuation coefficient among
these tissues due to its density and stiffness. Blood (A) has very low attenuation. Liver
(B) and fat (D) have moderate attenuation, but bone attenuates sound significantly
more.
Correct Answer: C
Question 8 Acoustic impedance is calculated as:
A. Density × propagation speed
B. Density ÷ propagation speed
C. Frequency × wavelength
D. Pressure ÷ volume
Rationale: A. Density × propagation speed [CORRECT]. Acoustic impedance (Z) = ρ
× c, where ρ is density and c is propagation speed. Option B inverts the relationship.
Option C defines speed, not impedance. Option D is unrelated to acoustic
impedance.
Correct Answer: A
Question 9 At an interface between soft tissue (Z ≈ 1.5 MRayl) and air (Z ≈ 0.0004
MRayl), the percentage of reflected intensity is approximately:
A. 0.1%
B. 1%
C. 50%
D. 99.9%
