Q79PP - University Physics with Modern Physics

Question

 Because the speed of ultrasound in bone is about twice the speed in soft tissue, the distance to a structure that lies beyond a bone can be measured incorrectly. If a beam passes through $\mathbf{4}\mathbf{ }\mathit{c}\mathit{m}$ of tissue, then $\mathbf{2}\mathbf{ }\mathit{c}\mathit{m}$ of bone, and then another $\mathbf{1}\mathbf{ }\mathit{c}\mathit{m}$ of tissue before echoing off a cyst and returning to the transducer, what is the difference between the true distance to the cyst and the distance that is measured by assuming the speed is always $\mathbf{1540}\mathbf{ }\mathit{m}\mathbf{/}\mathit{s}$? Compared with the measured distance, the structure is actually

(a) $\mathbf{1}\mathit{c}\mathit{m}$ farther

(b) $\mathbf{2}\mathbf{ }\mathit{c}\mathit{m}$ farther

(c) $\mathbf{1}\mathbf{ }\mathit{c}\mathit{m}$closer

(d) $\mathbf{2}\mathbf{ }\mathit{c}\mathit{m}$closer

Step-by-Step Solution

Answer

The correct option is a.

Step 1: Measure the total time

The beam goes $2cm$ into the bone and $5cm$ into the tissue with a velocity of $1540cm$ in the tissue and $3080m$ in the bone. The total time is given by

$$\begin{gathered} \text{t=}\frac{0.05m}{1540ms}+\frac{0.02m}{3080m/s} \\ \mathrm{t}=3.89\times {10}^{-5} \mathrm{s} \end{gathered}$$

The measured distance will be

$$\begin{gathered} d=vt \\ d=1540\times 3.89\times {10}^{-5} \\ d=0.06m \\ d=6cm \end{gathered}$$

Step 2: Calculate the difference

The total distance is $7cm$ while the measured distance is $6cm$. The difference will be

$$\begin{gathered} \Delta d=7-6 \\ \Delta d=1cm \end{gathered}$$

Hence, the correct option is $a$.