ARDMS SONOGRAPHY PRINCIPLES & INSTRUMENTATION (SPI) EXAM 2025 –
120-QUESTION PRACTICE EXAM WITH VERIFIED ANSWERS
PHYSICS PRINCIPLES
Which of the following describes the reciprocal relationship between period
and frequency?
A) As frequency increases, period increases.
B) As frequency increases, period decreases.
C) Period and frequency are unrelated.
D) As frequency decreases, period decreases.
Answer: B
1. Rationale: Period and frequency are inversely related. Period is the time it
takes for one cycle to occur, while frequency is the number of cycles that
occur in one second. Mathematically, Period (s) = 1 / Frequency (Hz).
Therefore, if frequency increases, the time for one cycle (period) must
decrease.
In soft tissue, the attenuation coefficient is approximately how many dB/cm for
every 1 MHz increase in frequency?
A) 0.5 dB/cm/MHz
B) 1.0 dB/cm/MHz
C) 2.0 dB/cm/MHz
D) 5.0 dB/cm/MHz
Answer: A
2. Rationale: The rule of thumb for the attenuation coefficient in soft tissue is
0.5 dB/cm for every 1 MHz of frequency. For example, a 5 MHz transducer
will have an attenuation coefficient of approximately 2.5 dB/cm in soft
tissue.
Which intensity measurement is most relevant when evaluating the potential
for thermal bioeffects during a pulsed ultrasound examination?
A) Spatial Peak, Pulse Average (SPPA)
B) Spatial Peak, Temporal Average (SPTA)
C) Spatial Average, Temporal Average (SATA)
D) Spatial Average, Pulse Average (SAPA)
Answer: B
3. Rationale: Spatial Peak, Temporal Average (SPTA) intensity is the average
intensity at the point in the beam where it is highest (spatial peak),
, averaged over the entire pulse cycle including the time the transducer is
"off" (temporal average). SPTA is the most widely accepted intensity
measurement for predicting thermal bioeffects because it accounts for both
where the beam is most intense and the overall exposure time.
Acoustic impedance is equal to:
A) Density multiplied by frequency
B) Density multiplied by propagation speed
C) Propagation speed divided by density
D) Frequency multiplied by wavelength
Answer: B
4. Rationale: Acoustic impedance (Z) is a characteristic of a medium and is
calculated by multiplying the density of the medium (kg/m³) by the
propagation speed of sound in that medium (m/s). Its unit is Rayls (kg/m²s).
If the acoustic impedance of medium 1 is 1.5 MRayls and the acoustic impedance
of medium 2 is 1.6 MRayls, what percentage of the incident sound intensity will
be reflected at the boundary?
A) Approximately 0.1%
B) Approximately 1.0%
C) Approximately 10%
D) Approximately 50%
Answer: A
5. Rationale: To calculate the intensity reflection coefficient, use the formula:
IRC = [(Z2 - Z1) / (Z2 + Z1)]². IRC = [(1.6 - 1.5) / (1.6 + 1.5)]² = (0..1)² =
(0.032)² = 0.001, or 0.1%. This demonstrates why sound easily transmits
between soft tissue interfaces with similar impedances.
According to Snell's Law, what must occur for refraction (bending of the sound
beam) to take place?
A) The angle of incidence must be 90 degrees.
B) The impedances of the two media must be identical.
C) The sound beam must strike the boundary at an oblique angle, and the
propagation speeds of the two media must be different.
D) The frequency of the sound must change as it crosses the boundary.
Answer: C
6. Rationale: Refraction is the bending of a sound beam. For refraction to
occur, two conditions must be met: 1) The beam must strike the interface at
, an oblique angle (not 90 degrees/normal incidence), and 2) The propagation
speeds of the two media on either side of the boundary must be different.
Frequency does not change when sound crosses a boundary.
What is the primary purpose of the range equation in diagnostic ultrasound?
A) To calculate the distance to a reflector based on the go-return time and the
speed of sound.
B) To calculate the frequency of the transducer needed to image a specific
depth.
C) To determine the amount of attenuation that will occur at a specific depth.
D) To calculate the angle of incidence required to avoid refraction.
Answer: A
7. Rationale: The range equation relates the distance (depth) of a reflector to
the time it takes for the pulse to travel to the reflector and back (go-return
time). The formula is d = (c * t) / 2. Because the time recorded by the
ultrasound machine is the "go-return" time, the speed of sound must be
divided by 2.
How is Pulse Repetition Frequency (PRF) related to depth of view (Maximum
imaging depth)?
A) PRF is directly proportional to maximum depth.
B) PRF is inversely proportional to maximum depth.
C) PRF and maximum depth are unrelated.
D) PRF is equal to maximum depth multiplied by 2.
Answer: B
8. Rationale: To wait for echoes to return from deeper structures, the
ultrasound system must increase the listening time (pulse repetition period).
Because PRF is the reciprocal of PRP (PRF = 1 / PRP), increasing the PRP to
image deeper structures results in a lower PRF. Therefore, PRF and
maximum imaging depth are inversely related.
A transducer operates at a frequency of 5 MHz. The pulse duration is 0.4
microseconds. The duty factor is:
A) 0.02%
B) 0.2%
C) 2.0%
D) 20%
Answer: C
, *Rationale: Duty factor is the fraction of time that the ultrasound system is
transmitting a pulse. It is calculated as Duty Factor = Pulse Duration / Pulse
Repetition Period. We know Pulse Duration (0.4 µs). To find PRP, we must first
find maximum depth or assume a standard PRF. Assuming a standard imaging
depth of ~15 cm (PRP = 200 µs), DF = 0. = 0.002, or 0.2%. However,
without PRP provided, let's look at standard duty factors. Diagnostic imaging
duty factors typically range from 0.1% to 1%. Wait, let's calculate using standard
formulas if PRP isn't given, it's a trick. Let me re-evaluate. If we use standard
PRF of 5 kHz (PRP = 200 µs), DF = 0.002 (0.2%). If PRF is 1 kHz (PRP = 1000 µs),
DF = 0.0004 (0.04%). Since none of these are options, let's check the math if PRP
was 20 µs (depth 1.5cm, unlikely). Let's assume the question implies a standard
depth. Let me correct the question to provide PRP. Self-correction: I will provide
PRP in the question to make it solvable.
Revised Question 9: A transducer operates at a frequency of 5 MHz. The pulse
duration is 0.4 microseconds and the pulse repetition period is 200
microseconds. The duty factor is:
Answer: B (0.2%)
9. Rationale: Duty Factor (DF) = Pulse Duration (PD) / Pulse Repetition Period
(PRP). DF = 0.4 µs / 200 µs = 0.002. To convert to a percentage, multiply by
100, which equals 0.2%. This is a typical duty factor for pulsed wave
ultrasound.
Spatial Pulse Length (SPL) is determined by:
A) Frequency and propagation speed
B) Number of cycles in the pulse and the wavelength
C) The amplitude of the pulse and the medium's attenuation
D) The PRF and the depth of view
Answer: B
10.Rationale: SPL is the physical length of a pulse in space. It is calculated by
multiplying the number of cycles in the pulse by the wavelength (SPL = n ×
λ). Since wavelength is determined by frequency and propagation speed (λ =
c / f), SPL is ultimately determined by the number of cycles, frequency, and
speed. However, the direct definition is cycles times wavelength.
Which of the following statements is true regarding the relationship between
stiffness, compressibility, and propagation speed?
A) As stiffness increases, propagation speed decreases.
B) As compressibility increases, propagation speed increases.
120-QUESTION PRACTICE EXAM WITH VERIFIED ANSWERS
PHYSICS PRINCIPLES
Which of the following describes the reciprocal relationship between period
and frequency?
A) As frequency increases, period increases.
B) As frequency increases, period decreases.
C) Period and frequency are unrelated.
D) As frequency decreases, period decreases.
Answer: B
1. Rationale: Period and frequency are inversely related. Period is the time it
takes for one cycle to occur, while frequency is the number of cycles that
occur in one second. Mathematically, Period (s) = 1 / Frequency (Hz).
Therefore, if frequency increases, the time for one cycle (period) must
decrease.
In soft tissue, the attenuation coefficient is approximately how many dB/cm for
every 1 MHz increase in frequency?
A) 0.5 dB/cm/MHz
B) 1.0 dB/cm/MHz
C) 2.0 dB/cm/MHz
D) 5.0 dB/cm/MHz
Answer: A
2. Rationale: The rule of thumb for the attenuation coefficient in soft tissue is
0.5 dB/cm for every 1 MHz of frequency. For example, a 5 MHz transducer
will have an attenuation coefficient of approximately 2.5 dB/cm in soft
tissue.
Which intensity measurement is most relevant when evaluating the potential
for thermal bioeffects during a pulsed ultrasound examination?
A) Spatial Peak, Pulse Average (SPPA)
B) Spatial Peak, Temporal Average (SPTA)
C) Spatial Average, Temporal Average (SATA)
D) Spatial Average, Pulse Average (SAPA)
Answer: B
3. Rationale: Spatial Peak, Temporal Average (SPTA) intensity is the average
intensity at the point in the beam where it is highest (spatial peak),
, averaged over the entire pulse cycle including the time the transducer is
"off" (temporal average). SPTA is the most widely accepted intensity
measurement for predicting thermal bioeffects because it accounts for both
where the beam is most intense and the overall exposure time.
Acoustic impedance is equal to:
A) Density multiplied by frequency
B) Density multiplied by propagation speed
C) Propagation speed divided by density
D) Frequency multiplied by wavelength
Answer: B
4. Rationale: Acoustic impedance (Z) is a characteristic of a medium and is
calculated by multiplying the density of the medium (kg/m³) by the
propagation speed of sound in that medium (m/s). Its unit is Rayls (kg/m²s).
If the acoustic impedance of medium 1 is 1.5 MRayls and the acoustic impedance
of medium 2 is 1.6 MRayls, what percentage of the incident sound intensity will
be reflected at the boundary?
A) Approximately 0.1%
B) Approximately 1.0%
C) Approximately 10%
D) Approximately 50%
Answer: A
5. Rationale: To calculate the intensity reflection coefficient, use the formula:
IRC = [(Z2 - Z1) / (Z2 + Z1)]². IRC = [(1.6 - 1.5) / (1.6 + 1.5)]² = (0..1)² =
(0.032)² = 0.001, or 0.1%. This demonstrates why sound easily transmits
between soft tissue interfaces with similar impedances.
According to Snell's Law, what must occur for refraction (bending of the sound
beam) to take place?
A) The angle of incidence must be 90 degrees.
B) The impedances of the two media must be identical.
C) The sound beam must strike the boundary at an oblique angle, and the
propagation speeds of the two media must be different.
D) The frequency of the sound must change as it crosses the boundary.
Answer: C
6. Rationale: Refraction is the bending of a sound beam. For refraction to
occur, two conditions must be met: 1) The beam must strike the interface at
, an oblique angle (not 90 degrees/normal incidence), and 2) The propagation
speeds of the two media on either side of the boundary must be different.
Frequency does not change when sound crosses a boundary.
What is the primary purpose of the range equation in diagnostic ultrasound?
A) To calculate the distance to a reflector based on the go-return time and the
speed of sound.
B) To calculate the frequency of the transducer needed to image a specific
depth.
C) To determine the amount of attenuation that will occur at a specific depth.
D) To calculate the angle of incidence required to avoid refraction.
Answer: A
7. Rationale: The range equation relates the distance (depth) of a reflector to
the time it takes for the pulse to travel to the reflector and back (go-return
time). The formula is d = (c * t) / 2. Because the time recorded by the
ultrasound machine is the "go-return" time, the speed of sound must be
divided by 2.
How is Pulse Repetition Frequency (PRF) related to depth of view (Maximum
imaging depth)?
A) PRF is directly proportional to maximum depth.
B) PRF is inversely proportional to maximum depth.
C) PRF and maximum depth are unrelated.
D) PRF is equal to maximum depth multiplied by 2.
Answer: B
8. Rationale: To wait for echoes to return from deeper structures, the
ultrasound system must increase the listening time (pulse repetition period).
Because PRF is the reciprocal of PRP (PRF = 1 / PRP), increasing the PRP to
image deeper structures results in a lower PRF. Therefore, PRF and
maximum imaging depth are inversely related.
A transducer operates at a frequency of 5 MHz. The pulse duration is 0.4
microseconds. The duty factor is:
A) 0.02%
B) 0.2%
C) 2.0%
D) 20%
Answer: C
, *Rationale: Duty factor is the fraction of time that the ultrasound system is
transmitting a pulse. It is calculated as Duty Factor = Pulse Duration / Pulse
Repetition Period. We know Pulse Duration (0.4 µs). To find PRP, we must first
find maximum depth or assume a standard PRF. Assuming a standard imaging
depth of ~15 cm (PRP = 200 µs), DF = 0. = 0.002, or 0.2%. However,
without PRP provided, let's look at standard duty factors. Diagnostic imaging
duty factors typically range from 0.1% to 1%. Wait, let's calculate using standard
formulas if PRP isn't given, it's a trick. Let me re-evaluate. If we use standard
PRF of 5 kHz (PRP = 200 µs), DF = 0.002 (0.2%). If PRF is 1 kHz (PRP = 1000 µs),
DF = 0.0004 (0.04%). Since none of these are options, let's check the math if PRP
was 20 µs (depth 1.5cm, unlikely). Let's assume the question implies a standard
depth. Let me correct the question to provide PRP. Self-correction: I will provide
PRP in the question to make it solvable.
Revised Question 9: A transducer operates at a frequency of 5 MHz. The pulse
duration is 0.4 microseconds and the pulse repetition period is 200
microseconds. The duty factor is:
Answer: B (0.2%)
9. Rationale: Duty Factor (DF) = Pulse Duration (PD) / Pulse Repetition Period
(PRP). DF = 0.4 µs / 200 µs = 0.002. To convert to a percentage, multiply by
100, which equals 0.2%. This is a typical duty factor for pulsed wave
ultrasound.
Spatial Pulse Length (SPL) is determined by:
A) Frequency and propagation speed
B) Number of cycles in the pulse and the wavelength
C) The amplitude of the pulse and the medium's attenuation
D) The PRF and the depth of view
Answer: B
10.Rationale: SPL is the physical length of a pulse in space. It is calculated by
multiplying the number of cycles in the pulse by the wavelength (SPL = n ×
λ). Since wavelength is determined by frequency and propagation speed (λ =
c / f), SPL is ultimately determined by the number of cycles, frequency, and
speed. However, the direct definition is cycles times wavelength.
Which of the following statements is true regarding the relationship between
stiffness, compressibility, and propagation speed?
A) As stiffness increases, propagation speed decreases.
B) As compressibility increases, propagation speed increases.
