What is the total attenuation of a 3.5 MHz pulse after passing through 2 cm of soft tissue?

Prepare for the SPI Ultrasound Test with comprehensive flashcards and multiple choice questions, each offering hints and explanations. Equip yourself for the exam!

To determine the total attenuation of a 3.5 MHz ultrasound pulse after passing through 2 cm of soft tissue, it's essential to use the known attenuation coefficients for soft tissue. The typical attenuation rate for soft tissue is approximately 0.5 dB/cm/MHz.

Given this information, the calculation for attenuation can be conducted as follows:

  1. Establish the attenuation per cm using the frequency:
  • At 3.5 MHz, the soft tissue attenuation would be 0.5 dB/cm/MHz × 3.5 MHz = 1.75 dB per cm.
  1. Next, multiply the attenuation per cm by the total distance (2 cm):
  • 1.75 dB/cm × 2 cm = 3.5 dB.

This means that after the pulse has traveled through 2 cm of soft tissue, it experiences a total attenuation of 3.5 dB. This result aligns with the answer provided. Understanding these calculations and the fundamental relationship between frequency, distance, and attenuation helps reinforce the principles of ultrasound physics related to tissue interaction.

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