Tsunamis, in the deep ocean, travel at around 800 kilometers per hour.

The speed of sound under water is about 5300 kilometers per hour.

Both of these waves are pressure waves, operating in the same medium. Why is one so much faster than the other?


1 Answer 1


Tsunamis and sound waves are different types of wave - one is a transverse wave and the other is a longitudinal one. Let's look at the factors that influence the speed of each one.

Tsunami - transverse wave in shallow water

A transverse wave is one of the type that we think of from day to day - where the direction of oscillation is perpendicular to the direction of travel. The speed that a transverse wave travels at depends on different factors depending on the depth of the water. For this purpose, "shallow water" is usually defined as existing where depth < wavelength/20. The wavelength of a tsunami is very large - of the order of hundreds of kilometres - so for a tsunami, any part of the world's oceans counts as "shallow water".

In shallow water, the speed of a transverse wave can be described by,

$$V = \sqrt{gD}$$

where $V$ is the wave's speed, $D$ is the depth, and $g$ is the acceleration due to gravity (9.81 m/s2). In the case of a tsunami in the deep ocean, then, if we assume a depth of 4 km we can estimate a speed of 198 m/s, or 713 kph. That's a back-of-an-envelope calculation, but it's sufficiently similar to the 800kph that you quoted in the question that I'm happy with it.

Sound - longitudinal wave

In a longitudinal wave, the direction of the oscillation is parallel to the direction of travel - i.e. it's an oscillation in the density of the material. We don't see many of these in everyday life, but one good example is the wave that moves down a slinky if you jerk the ends towards or away from each other.

Picture of a slinky showing a longitudinal wave (Image source)

Sound, in a liquid or a gas, is an example of a longitudinal wave. The speed of a longitudinal wave depends on the stiffness and the density of the material that it travels through, in the following way:

$$c = \sqrt{\frac{K}{\rho}}$$

where $c$ is the speed of the wave, $K$ is the bulk modulus of the fluid, and $\rho$ is its density. Note that there is no dependence on the depth of the fluid (in this case the sea) - sound one metre below the surface of a salt water swimming pool would move at roughly the same speed as sound one meter below the surface of an ocean.


Sound waves and tsunami waves propagate through different mechanisms, and thus different things influence their speeds.

  • 1
    $\begingroup$ Nice answer. It would be excellent if there was some more explanation of why the two equations are used - maybe not the full derivation, but something indicating how they were discovered, perhaps? $\endgroup$
    – naught101
    Commented Apr 16, 2014 at 12:10
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    $\begingroup$ This could be further expanded, if not corrected, with the fact that in a tsunami there is a net movement of water, instead of, as with normal waves, only net transfer of energy. Soundwaves then again are longitudinal pressure waves, and in that case indeed different wave velocities apply. $\endgroup$
    – hugovdberg
    Commented Apr 17, 2014 at 8:53
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    $\begingroup$ @hugovdberg: I'd be interested to see that. I don't understand how there can be a net movement of water, at least over the long term (weeks+). Isn't it just a matter of scale? $\endgroup$
    – naught101
    Commented Nov 6, 2014 at 2:06
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    $\begingroup$ In short, tsunamis are usualy caused by a net movement of the ocean floor, which typically rises several meters over a large area. That displaced water bulges over that area and then flows sideways as a very shallow wave (in the order of centimetres, hardly noticable over the normal wind wave s) which only increases in height as soon as that wall of water hits the continental shelf. But since the seafloor does not return to its initial position neither will the water. $\endgroup$
    – hugovdberg
    Commented Nov 6, 2014 at 7:00
  • 1
    $\begingroup$ Related discussion on Physics SE. $\endgroup$
    – Matt Hall
    Commented May 12, 2015 at 17:17

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