During a fog, a navigator receives two signals simultaneously sent by a station on the coast, one through the air and
the other through the water. Between the receptions of the two sounds, a time interval t=5 seconds elapses. Under the
conditions of the experiment, the speed of sound in the air is 341 m/s and 1504 m/s in the water. Find the distance
x between the boat and the station emitting the signals.
Problem data
- speed of sound in air: var = 341 m/s;
- speed of sound in water: vag = 1504 m/s;
- the time interval between receptions: Δ t = 5 s.
Problem diagram
As the sound wave that spreads through the water has greater speed it arrives before to the boat. If
t is the
wave propagation time in the water, the wave propagation time in the air is
t+Δ
t (it will be
the sum of the time
t of propagation in the water with the delay Δ
tthat it has to be slower).
We chose a reference frame oriented to the right with origin in the sound emission position. The initial position of
the sound waves will be
S0ar =
S0ag = 0, the position of the boat is
x,
the final position where the waves should arrive
Sar =
Sag = x.
Solution
The equation of displacement as a function of time is given by
\[ \bbox[#99CCFF,10px]
{S=S_{0}+\mathit{vt}} \tag{I}
\]
Using the expression (I) for the wave propagating through water, we have
\[
S_{\text{w}}=S_{0\text{w}}+v_{\text{w}}t
\]
substituting the problem data
\[
\begin{align}
x=0+1504t\\
x=1504t \quad \tag{II}
\end{align}
\]
using the expression (I) for the wave propagating through the air, we have
\[
S_{\text{a}}=S_{0\text{a}}+v_{\text{a}}t
\]
substituting the problem data
\[
\begin{align}
x=0+341\;(\;t+\Delta t\;)\\
x=341\;(\;t+5\;) \quad \tag{III}
\end{align}
\]
From the expression (II) we have the time t
\[
\begin{align}
t=\frac{x}{1504} \tag{IV}
\end{align}
\]
substituting the expression (IV) in (III), we have
\[
x=341\;\left(\;\frac{x}{1504}+5\;\right)\\
x=\frac{341}{1504}x+341 \times 5\\
x=\frac{341}{1504}x+1705\\
x-\frac{341}{1504}x=1705
\]
the
Least Common Multiple (
LCM) between 1 and 1504 is 1504
\[
\frac{1504x-341x}{1504}=1705\\
1163x=1705 \times 1504\\
1163x=2564320\\
x=\frac{2564320}{1163}
\]
\[ \bbox[#FFCCCC,10px]
{x\simeq 2205\;\text{m}}
\]