Kvant Physics Problem 179
A submarine moves vertically downward with constant speed $u$ in still water where sound propagates with speed $V$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m30s
Source on kvant.digital
Problem
In the journal it was mistakenly published under number 181.
A submarine, while descending vertically, emits short sonar pulses of duration $\tau_0$ toward the seabed. The duration of the reflected signals, measured by the hydroacoustician on the submarine, 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 with constant speed $u$ in still water where sound propagates with speed $V$. The submarine emits short sonar pulses of duration $\tau_0$, which propagate to a horizontal seabed, reflect, and return to the submarine. The received echo has duration $\tau$ as measured on the submarine.
The unknown is the descent speed $u$ with units $\mathrm{m/s}$. The known quantities are $\tau_0$ and $\tau$ in seconds, and $V$ in $\mathrm{m/s}$.
The water is treated as a uniform stationary medium. The seabed is perfectly rigid and stationary, producing specular reflection without changing frequency. The motion is strictly collinear with wave propagation. The speed $u$ is assumed constant during the emission and reception of each pulse. Dissipative attenuation and dispersion of sound are neglected.
Physical Principles
Sound propagation in a moving source frame relative to a stationary medium produces a change in wavelength determined by the source motion. For a source moving toward the propagation direction with speed $u$, successive wavefronts are emitted from progressively closer positions, giving the wavelength in the medium as
$$\lambda = (V - u)\tau_0.$$
Reflection at a stationary boundary preserves frequency and therefore preserves the temporal spacing between wavefronts in the medium.
For an observer moving with speed $u$ toward an incoming wave of wavelength $\lambda$, the observed frequency is increased according to the classical Doppler relation
$$f_{\text{obs}} = \frac{V + u}{\lambda}.$$
The observed period is the inverse of the observed frequency,
$$\tau = \frac{1}{f_{\text{obs}}}.$$
Derivation
The submarine emits pulses with intrinsic period $\tau_0$. During emission, the submarine moves downward toward the seabed, so each successive pulse is emitted from a closer position to the reflector. In the water frame, the spatial separation between successive wavefronts equals the distance sound travels in one emission period minus the distance the source moves in the same time, giving the wavelength
$$\lambda = V\tau_0 - u\tau_0 = (V - u)\tau_0.$$
After reflection at the seabed, the wave propagates upward toward the submarine without change in frequency, so the same wavelength $\lambda$ characterizes the returning wave in the water frame.
During reception, the submarine moves downward toward the incoming upward-propagating wavefronts. In time $\Delta t$, wavefronts approach the submarine by distance $V\Delta t$ while the submarine moves upward relative to the wave by an additional effective closing speed $u\Delta t$. The relative rate at which wavefronts meet the submarine corresponds to an effective propagation speed $V + u$, giving the observed frequency
$$f = \frac{V + u}{\lambda}.$$
Substituting the expression for $\lambda$,
$$f = \frac{V + u}{(V - u)\tau_0}.$$
The measured duration of the received pulse is the inverse frequency,
$$\tau = \frac{1}{f} = \tau_0 \frac{V - u}{V + u}.$$
Solving this relation for $u$ begins by multiplying both sides by $V + u$,
$$\tau (V + u) = \tau_0 (V - u).$$
Expanding terms,
$$\tau V + \tau u = \tau_0 V - \tau_0 u.$$
Collecting terms containing $u$ on one side and constant terms on the other,
$$\tau u + \tau_0 u = \tau_0 V - \tau V.$$
Factoring $u$,
$$u(\tau + \tau_0) = V(\tau_0 - \tau).$$
Dividing by $\tau + \tau_0$,
$$u = V \frac{\tau_0 - \tau}{\tau_0 + \tau}.$$
Result
The descent speed of the submarine is
$$u = V \frac{\tau_0 - \tau}{\tau_0 + \tau}.$$
Substituting quantities with units,
$$u = V \frac{\tau_0 - \tau}{\tau_0 + \tau} = \left(\mathrm{m/s}\right)\frac{\mathrm{s} - \mathrm{s}}{\mathrm{s} + \mathrm{s}} = \mathrm{m/s}.$$
The final boxed result is
$$\boxed{u = V \frac{\tau_0 - \tau}{\tau_0 + \tau}}.$$
Sanity Checks
When $u = 0$, the expression gives $\tau = \tau_0$, which matches the absence of motion since no Doppler deformation occurs.
As $u \to V$, the numerator in $\tau = \tau_0 \frac{V - u}{V + u}$ tends to zero while the denominator remains finite, producing $\tau \to 0$, consistent with extreme compression of the returned pulse as the submarine approaches the sound speed in water.
Dimensional consistency holds since the fraction $\frac{\tau_0 - \tau}{\tau_0 + \tau}$ is dimensionless, leaving $u$ with units of $V$, namely $\mathrm{m/s}$.
The most sensitive step to sign errors arises in the wavelength formation $\lambda = (V - u)\tau_0$ and in the observer factor $V + u$, where reversing either relative motion direction would invert the Doppler ratio and incorrectly predict pulse stretching instead of compression.