Question

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.12 ms. At what depth (in cm) did this reflection occur? (Use any necessary data found in this table.)

Answer 1

We are asked to determine the distance traveled by a sound wave when it is reflected in fat tissue. To do that we will use the fact that the velocity is the quotient between the distance and the time:

$v=\frac{d}{t}$

Where:

$\begin{gathered} v=\text{ velocity of sound in fat tissue} \\ d=\text{ distance} \\ t=\text{ time} \end{gathered}$

Now we multiply both sides by "t":

$vt=d$

Since the time is equivalent to 2 times the depth that the sound traveled then we have:

$vt_d=2d$

Where:

$t_d=\text{ time delay}$

Now, we substitute the values. We use the value of the velocity of sound found in the table:

$(1450\frac{m}{s})(0.12\times10^{-3}s)=2d$

Now, we divide both sides by 2:

$\frac{(1450\times\frac{m}{s})(0.12e-3\cdot10^{-3}s)}{2}=d$

Solving the operations:

$0.087m=d$

Now, we convert to cm by multiplying by 100:

$d=0.087m\times\frac{100cm}{1m}=8.7cm$

Therefore, the depth is 8.7 cm.