Question

Two transducers send ultrasound pulses into soft tissue. One transducer emits sound with a 4 MHz frequency, and the other produces sound at a 6 MHz frequency. Each pulse contains 4 cycles. Which has a greater spatial pulse length?

A. The 6 MHz pulse
B. The 4 MHz pulse
C. They are the same
D. Cannot be determined

Answer 1

The spatial pulse length for a lower frequency pulse is greater, given the same number of cycles. Hence, the 4 MHz transducer with a longer wavelength than the 6 MHz transducer produces a greater spatial pulse length.

Step-by-step explanation:

When considering ultrasound pulses emitted by transducers, the spatial pulse length is determined by both the frequency of the sound wave and the number of cycles in each pulse. The spatial pulse length is calculated by multiplying the wavelength of the sound by the number of cycles in the pulse. Since the wavelength is inversely proportional to the frequency (given by λ = v/f where λ is the wavelength, v is the speed of sound, and f is the frequency), a lower frequency ultrasound will have a longer wavelength. Therefore, for the same number of cycles, a lower frequency pulse will have a greater spatial pulse length.

The speed of sound in soft tissue is approximately 1540 m/s. Using this speed:

For the 4 MHz transducer: Wavelength λ = 1540 m/s / (4×106 Hz) = 0.385 mm. Spatial pulse length = 0.385 mm × 4 cycles = 1.54 mm.

For the 6 MHz transducer: Wavelength λ = 1540 m/s / (6×106 Hz) = 0.257 mm. Spatial pulse length = 0.257 mm × 4 cycles = 1.028 mm.

Since 1.54 mm is greater than 1.028 mm, the 4 MHz pulse has a greater spatial pulse length. Therefore, the correct answer is B. The 4 MHz pulse.