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=(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(m)/(s))(0.12*10^(-3)s)=2d`

Now, we divide both sides by 2:

`((1450*(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*(100cm)/(1m)=8.7cm`

Therefore, the depth is 8.7 cm.