Question

A typical ultrasound transducer used for medical diagnosis produces a beam of ultrasound with a frequency of 1.0 MHz. The beam travels from the transducer through tissue and partially reflects when it encounters different structures in the tissue. The same transducer that produces the ultrasound also detects the reflections. The transducer emits a short pulse of ultrasound and waits to receive the reflected echoes before emitting the next pulse. By measuring the time between the initial pulse and the arrival of the reflected signal, we can use the speed of ultrasound in tissue, 1540 m/s, to determine the distance from the transducer to the structure that produced the reflection. As the ultrasound beam passes through tissue, the beam is attenuated through absorption. Thus deeper structures return weaker echoes. A typical attenuation in tissue is -100 dB/m MHz; in bone it is -500 dB/m MHz. In determining attenuation, we take the reference intensity to be the intensity produced by the transducer.In some applications of ultrasound, such as its use on cranial tissues, large reflections from the surrounding bones can produce standing waves. This is of concern because the large pressure amplitude in an antinode can damage tissues. For a frequency of 1.0 MHz, what is the distance between antinodes in tissue? a) 0. 38 mm; b) 0.75 mm; c) 1,5 mm; d) 3.0 mm.

Answer 1

frequency of the ultrasound given as

`f = 1.0 MHz`

speed of the wave is

`v = 1540 m/s`

Now the wavelength is given as

`wavelength = (speed)/(frequency)`

`\lambda = (1540)/(1* 10^6)`

`\lambda = 1.5 mm`

now we know that distance between two consecutive antinodes is always half of wavelength

so we have

d = 0.75 mm