Question

Echo times are measured by diagnostic ultrasound scanners to determine distances to reflecting surfaces in a patient. What is the difference in echo times (answer in milliseconds) for muscle tissues that are 3.414 and 3.6266 cm beneath the surface? (This difference is the minimum resolving time for the scanner to see details as small as 0.100 cm, or 1.00 mm. Discrimination of smaller time differences is needed to see smaller details.); Note: According to table 17.5 the speed of sound in muscle tissue is 1590 m/s.

Answer 1

Let 1 m = 100 cm

Let d1 = depth of muscle tissue 1 = 3.414 cm = (3.414 x 0.01) m = 0.03414 m

Let d2 = depth of muscle tissue 2 = 3.6266 cm = (3.6266 x 0.01) m = 0.036266 m

Let v = speed of sound in muscle tissue = 1590 m/s

Let t1 = time of return journey of ultrasound signal in tissue 1 = 2 x d1 / v = 2 x 0.03414 m / 1590 m/s = 4.29434 x 10^-5 s

Let t2 = time of return journey of ultrasound signal in tissue 2 = 2 x d2 / v = 2 x 0.036266 m / 1590 m/s = 4.56176 x 10^-5 s

Let dt = difference in echo times for tissues 1 and 2 = t2 - t1 = (4.56176 x 10^-5 s) - (4.29434 x 10^-5 s) = 2.6742 x 10^-6 s

dt - difference in echo times of ultrasound signal for tissues 1 and 2 - is given by:

dt = t2 - t1 = (4.56176 x 10^-5 s) - (4.29434 x 10^-5 s) = 2.6742 x 10^-6 s

dt = 2.6742 x 10^-6 s

dt = 2.674 x 10^-6 s ---------------------------------------------------------------(answer)

Answer: Difference in echo times of ultrasound signal for tissues 1 and 2 is given by, dt = 2.674 x 10^-6 s

(Answer is rounded to four significant digits.)