Kvant Physics Problem 44
The physical system consists of two dolphins moving through water along the same straight line toward one another.
Verified: yes
Verdicts: PASS + PASS
Solve time: 5m46s
Source on kvant.digital
Problem
F44. Two dolphins are swimming toward each other. One of them emits a sound of frequency $\omega$. What frequency sound will the other dolphin hear if the dolphins' speeds relative to the water are equal and are $v$? The speed of sound in water is $u$.
All-Union Physics Olympiad, 1968.
Setup and Assumptions
The physical system consists of two dolphins moving through water along the same straight line toward one another. One dolphin acts as a source of sound and emits a wave of frequency $\omega$. The second dolphin acts as an observer and receives the sound.
The given quantities are the emitted frequency $\omega$ in hertz, the speed of each dolphin relative to the water $v$ in $\mathrm{m/s}$, and the speed of sound in water $u$ in $\mathrm{m/s}$.
The unknown is the frequency $\omega'$ heard by the receiving dolphin.
The water is taken as the medium in which sound propagates. The speed of sound relative to the water is constant and equal to $u$. The dolphins move with constant velocities. Sound propagation is assumed to be one-dimensional along the line joining the dolphins. Effects of water currents, acceleration, attenuation, and reflections are neglected.
Physical Principles
The solution is based on the Doppler effect for waves propagating in a material medium.
When a source moves toward a stationary medium observer with speed $v_s$, the wavelength in front of the source becomes
$$\lambda=\frac{u-v_s}{f},$$
where $f$ is the emitted frequency and $u$ is the wave speed in the medium.
When an observer moves toward incoming waves with speed $v_o$, the frequency received is
$$f'=\frac{u+v_o}{\lambda}.$$
These relations must be applied successively because both the source and the observer move relative to the water.
Derivation
Let the emitting dolphin have speed
$$v_s=v,$$
and let the receiving dolphin have speed
$$v_o=v.$$
The source emits waves with frequency
$$f=\omega.$$
During one period
$$T=\frac{1}{\omega},$$
the source advances a distance
$$vT=\frac{v}{\omega}.$$
Meanwhile the previously emitted wave crest moves a distance
$$uT=\frac{u}{\omega}.$$
Hence the distance between successive crests in front of the source, which is the wavelength propagating toward the second dolphin, equals
$$\lambda=uT-vT =\frac{u-v}{\omega}.$$
The second dolphin swims toward these incoming wave crests with speed $v$. Relative to the observer, the crests approach with speed
$$u+v.$$
The frequency heard by the observer is the rate at which crests arrive:
$$\omega'=\frac{u+v}{\lambda}.$$
Substituting the expression for $\lambda$ gives
$$\omega' =\frac{u+v}{(u-v)/\omega} =\omega,\frac{u+v}{u-v}.$$
This is the Doppler-shifted frequency.
Result
The frequency heard by the second dolphin is
$$\boxed{\omega'=\omega,\frac{u+v}{u-v}}.$$
Since no numerical values are provided in the problem statement, the result remains symbolic. The received frequency in hertz is
$$\boxed{\omega'=\omega,\frac{u+v}{u-v}\ \text{Hz}}$$
when $\omega$ is expressed in hertz and $u$ and $v$ are expressed in the same velocity units.
Sanity Checks
The factor
$$\frac{u+v}{u-v}$$
is dimensionless, so $\omega'$ has the same units as $\omega$, which confirms dimensional consistency.
If the dolphins are stationary, $v=0$, then
$$\omega'=\omega,$$
which matches physical expectations.
For small speeds compared with the speed of sound, $v\ll u$,
$$\omega' =\omega\frac{1+v/u}{1-v/u} \approx\omega\left(1+\frac{2v}{u}\right).$$
The shift is proportional to the closing speed $2v$, which is reasonable because both dolphins contribute equally to the Doppler effect.
As $v\to u$, the denominator approaches zero and the predicted frequency grows without bound. This signals the well-known singular behavior of the classical Doppler formula as the source speed approaches the wave speed in the medium.
The step most vulnerable to a sign error is the determination of the wavelength in front of the moving source. For approaching motion the wavelength is
$$\lambda=\frac{u-v}{\omega},$$
not $\frac{u+v}{\omega}$. Reversing this sign would incorrectly predict a lower received frequency instead of the higher frequency expected when source and observer move toward one another.