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College Physics (Urone)
Found in: Page 631

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Short Answer

(a) Calculate the minimum frequency of ultrasound that will allow you to see details as small as 0.250mm in human tissue. (b) What is the effective depth to which this sound is effective as a diagnostic probe?

(a) Minimum frequency of ultrasound will be \[6.2 \times {10^{6{\rm{ }}}}{\rm{ }}Hz\].

(b) Effective depth is 0.125m.

See the step by step solution

Step by Step Solution

Step 1: The frequency of ultrasound:

Ultrasound frequency is defined as the number of ultrasound waves per second.

The product of frequency and wavelength is the speed of the wave.

Step 2: (a) The minimum frequency of ultrasound

The minimum frequency is given as,

\[v = {f_{\min }}\lambda \]

Here, v the speed of sound in tissue, \[{f_{\min }}\] is the minimum frequency, and \[\lambda \] is the wavelength of an ultrasound wave that would be equal to the small traveling distance of wave in human tissue.

Consider the known data below.

The speed of tissue, \[v = 1540{\rm{ }}m{\rm{ }}{s^{ - 1}}\]

The wavelength, \[\lambda = 0.250{\rm{ }}mm = 2.5 \times {10^{ - 4}}{\rm{ }}m\]

Rearrange equation (1) for frequency as below and substitute known values.

\begin{aligned}{f_{\min }} & = \frac{v}{\lambda }\\ &= \frac{{1540}}{{2.5 \times {{10}^{ - 4}}}}\\ &= 6.2 \times {10^6}{\rm{ }}Hz \end{aligned}

Step 3: (b) Effective depth

The effective depth of sound wave to which this sound is effective as a diagnostic probe is given by,

\begin{aligned}{d_e} & = 500\lambda \\ &= 500 \times 2.5 \times {10^{ - 4}}\\ &= 0.125{\rm{ }}m\end{aligned}

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