Q80PP - University Physics with Modern Physics

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 $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{MHz}$ , what is the distance between antinodes in tissue?

Step-by-Step Solution

Answer

The correct option is  $b$.

Step 1: Given Data

The given frequency is  $1.00\text{\hspace{0.17em}MHz}$.

Step 2: Relation between distance and wavelength

The distance between nodes is related to the wavelength as

 $\mathbf{d}\mathbf{=}\frac{\mathrm{n\lambda }}{2}\mathbf{=}\frac{\mathrm{nv}}{2f}$

Step 3: Calculate the distance

Put the values of  $v$,  $f$ and  $n=1$ to get  $d$-

$\begin{array}{c}d=\frac{nv}{2f}\\ =\frac{1\times 1540\text{\hspace{0.17em}m}/\text{s}}{2\times {10}^6\text{\hspace{0.17em}Hz}}\\ =0.77\times {10}^{-3}\text{m}\\ =0.77\text{\hspace{0.17em}mm}\end{array}$

 

The result is   $d=0.77\text{\hspace{0.17em}mm}$.

Hence, the distance between antinodes is approximately  $0.75 mm$.