Kvant Physics Problem 179

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m11s
Source on kvant.digital

Problem

It was incorrectly published in the journal under problem number 181.

A submarine, descending vertically, emits short sonar sound pulses directed toward the seabed, with duration $\tau_0$. The duration of the reflected signals measured on board the submarine by the hydroacoustics operator is $\tau$. What is the speed of descent of the submarine?

The speed of sound in water is $V$. The seabed is horizontal.

Setup and Assumptions

A submarine moves vertically downward in a homogeneous, motionless fluid where sound propagates with constant speed $V$ relative to the water. The seabed is a fixed, perfectly reflecting horizontal surface. The submarine emits a short sonar pulse of duration $\tau_0$ as measured in the submarine frame. The reflected pulse returns to the same receiver on board, and its measured duration is $\tau$.

The unknown is the constant descent speed $u$ of the submarine, directed downward.

The analysis assumes rectilinear uniform motion of the submarine during the emission and reception of the pulse, instantaneous reflection at the seabed, and negligible absorption and dispersion of sound. The pulse duration is assumed short enough that $u$ remains constant over the entire emission and reception process.

Physical Principles

Sound pulses propagate in the water with speed $V$ relative to the medium, independent of the motion of the source and receiver.

The travel time of a signal is determined by the ratio of the path length in the water frame to the wave speed $V$.

The motion of the submarine during propagation changes the effective emission and reception geometry, so the total delay is obtained by summing emission time, downward propagation time, and upward propagation time computed in the same inertial frame of the water.

The duration of the received pulse is the difference between arrival times of the leading and trailing edges of the emitted pulse.

Derivation

Let the downward direction be positive. At emission time $t$, the submarine is at depth $x(t)$, and the distance to the seabed is $d(t)$. Since the submarine descends uniformly with speed $u$, the distance decreases as

$$d(t) = d_0 - ut.$$

The leading edge of a pulse emitted at time $t$ reaches the seabed after a time $t_1$ determined by the condition that sound travels the distance $d(t)$ at speed $V$:

$$V t_1 = d(t),$$

so

$$t_1 = \frac{d(t)}{V}.$$

At the reflection moment, the submarine has moved further downward during time $t_1$, so the distance between submarine and seabed becomes

$$d_{\mathrm{ref}}(t) = d(t) - u t_1 = d(t)\left(1 - \frac{u}{V}\right).$$

The reflected pulse propagates upward toward a receiver that continues moving downward. The relative closing speed between the upward-moving sound and the downward-moving submarine is $V+u$, so the return time $t_2$ satisfies

$$(V+u)t_2 = d_{\mathrm{ref}}(t),$$

which gives

$$t_2 = \frac{d(t)\left(1 - \frac{u}{V}\right)}{V+u}.$$

The arrival time of the reflected signal corresponding to emission time $t$ is

$$T(t) = t + t_1 + t_2 = t + \frac{d(t)}{V} + \frac{d(t)\left(1 - \frac{u}{V}\right)}{V+u}.$$

Since $d(t)=d_0-ut$ is linear in $t$, the function $T(t)$ is also linear in $t$, so the duration of the received pulse is obtained by multiplying the emitted duration by the coefficient $dT/dt$.

Differentiating, using $d'(t)=-u$, yields

$$\frac{dT}{dt} = 1 - \frac{u}{V} - \frac{u\left(1 - \frac{u}{V}\right)}{V+u}.$$

The received duration satisfies

$$\tau = \tau_0 \frac{dT}{dt}.$$

Introduce the ratio

$$r = \frac{\tau}{\tau_0}.$$

Then

$$r = 1 - \frac{u}{V} - \frac{u\left(1 - \frac{u}{V}\right)}{V+u}.$$

Multiplying by $V(V+u)$ eliminates denominators:

$$rV(V+u) = V(V+u) - u(V+u) - uV + u^2.$$

Expanding both sides and simplifying gives

$$rV(V+u) = V(V-u).$$

Dividing by $V$ leads to

$$r(V+u) = V - u.$$

Solving for $u$,

$$rV + ru = V - u,$$

$$u(r+1) = V(1-r),$$

so

$$u = V\frac{1-r}{1+r}.$$

Restoring $r=\tau/\tau_0$,

$$u = V\frac{\tau_0 - \tau}{\tau_0 + \tau}.$$

Result

The descent speed of the submarine is

$$u = V\frac{\tau_0 - \tau}{\tau_0 + \tau}.$$

Substituting the given quantities in consistent units,

$$\boxed{u = V\frac{\tau_0 - \tau}{\tau_0 + \tau}\ \text{m/s}}.$$

Sanity Checks

The expression is dimensionally consistent because it is a dimensionless ratio multiplied by the sound speed $V$, giving units of velocity.

In the limiting case $u=0$, the submarine is stationary, so $\tau=\tau_0$, and the formula gives $u=0$ as required.

If $u$ increases, the received pulse shortens, so $\tau<\tau_0$, making the numerator positive and yielding $u>0$, consistent with downward motion.

If $u$ approaches $V$, the received duration tends to zero since the reflected signal is rapidly overtaken by the descending receiver, and the formula gives $\tau \to 0$ and $u \to V$, matching the physical limit of sound propagation relative to the moving submarine.

The most sensitive step is the conversion of the round-trip propagation times into the single factor $dT/dt$, since any sign error in the motion during either the downward or upward leg would incorrectly change the final fractional dependence on $u/V$.