Question

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

Answer:

b. 0.75 mm

Explanation:

The distance between antinodes d is half the wavelength $\lambda$. We can obtain the wavelength with the formula $v=\lambda f$, where f is the frequency given ($f=1MHz=1\times10^6Hz$) and v is the speed of sound in body tissues (v=1540m/s), so putting all together we have:

$d=\frac{\lambda}{2}=\frac{v}{2f}=\frac{1540m/s}{2(1\times10^6Hz)}=0.00077m=0.77mm$

which is very close to the 0.75mm option.