ARDMS SPI Practice Questions (2026) — Free Sample Questions with Answers

Answer: B

0.15 mm: This is half the correct value, as if 1.54 mm/µs had been divided by 10 MHz.

0.31 mm: Correct. λ = c / f = 1.54 mm/µs ÷ 5 MHz = 0.308 mm. A handy rule: λ (mm) = 1.54 ÷ f (MHz) in soft tissue.

0.62 mm: This doubles the correct value, as if a 2.5-MHz transducer were used.

3.08 mm: This is a factor-of-10 unit error; at 5 MHz the wavelength in soft tissue is about 0.3 mm.

Wavelength = propagation speed ÷ frequency. In soft tissue λ (mm) = 1.54 / f (MHz), so higher frequency gives a shorter wavelength.

Answer: A

Heat above the Curie point depolarizes the crystal: Correct. Heating PZT above its Curie temperature destroys its polarization, so the element loses its piezoelectric properties permanently.

Heat raises the propagation speed of the lens: Heat damage is not about lens speed; the main hazard is depolarization of the piezoelectric element.

Heat permanently magnetizes the matching layer: Matching layers are not magnetic; transducers work by piezoelectricity, not magnetism.

Heat lowers the impedance of the backing block: Changes in backing impedance are not the reason; the critical risk is loss of piezoelectric function.

Never heat-sterilize transducers: exceeding the Curie temperature depolarizes the piezoelectric element. Follow the manufacturer's approved disinfection methods instead.

Answer: C

Increase the overall receiver gain: Gain brightens echoes but does not narrow the beam, so the structures remain merged.

Increase the pulse repetition frequency: PRF affects temporal resolution and maximum depth, not beam width.

Move the focal zone to a depth of 8 cm: Correct. Lateral resolution equals beam width, which is narrowest at the focus; placing the focus at the structures improves side-by-side resolution.

Lengthen the spatial pulse length: A longer pulse worsens axial resolution and has no benefit for lateral resolution.

Lateral resolution = beam width: best at the focal zone. Set the focus at or just below the region of interest.

Answer: A

The Nyquist limit is 2.5 kHz, so aliasing occurs: Correct. Nyquist limit = PRF / 2 = 2.5 kHz; a 3.2-kHz shift exceeds it, so the spectrum wraps around (aliases).

The Nyquist limit is 5 kHz, so no aliasing occurs: The Nyquist limit is half the PRF, not equal to it.

The Nyquist limit is 10 kHz, so no aliasing occurs: 10 kHz doubles the PRF; the limit is PRF ÷ 2.

The Nyquist limit is 1.6 kHz, so aliasing occurs: 1.6 kHz is half the Doppler shift; the Nyquist limit depends on the PRF, not on the shift.

Nyquist limit = PRF ÷ 2. Aliasing occurs in pulsed Doppler when the Doppler shift exceeds the Nyquist limit.

Answer: C

0.65 kHz: 0.65 kHz omits the factor of 2 in the Doppler equation (the shift occurs on both transmission and reception).

2.25 kHz: 2.25 kHz uses cos 30° (0.866) instead of cos 60° (0.5).

1.3 kHz: Correct. Δf = 2 f₀ v cosθ / c = 2 × 4,000,000 × 0.5 m/s × 0.5 / 1,540 m/s ≈ 1,299 Hz ≈ 1.3 kHz.

2.6 kHz: 2.6 kHz uses cos 0° (1.0), as if the beam were parallel to flow.

Doppler shift = 2 f₀ v cosθ / c. The shift is greatest parallel to flow (cos 0° = 1) and zero at 90°; keep units consistent (Hz, m/s).

Answer: B

Increase the transmit output power: Raising output also brightens the image but increases patient exposure; under ALARA it is tried only after receiver gain.

Increase the receiver gain or TGC: Correct. Receiver gain amplifies returning echoes without increasing the acoustic energy delivered to the patient.

Increase the number of focal zones: More focal zones improve lateral resolution over depth but do not correct overall darkness and lower the frame rate.

Increase the dwell time on the area: Longer dwell time increases exposure without improving brightness; ALARA calls for minimizing scan time.

ALARA: use the lowest output and shortest exposure that give a diagnostic image, so increase receiver gain before output power.