A rock rests on a concrete sidewalk. An earthquake strikes, making the ground mo

Question

A rock rests on a concrete sidewalk. An
earthquake strikes, making the ground move vertically
in simple harmonic motion with a constant frequency of
2.40 Hz and with gradually increasing amplitude. (a) With
what amplitude does the ground vibrate when the rock
begins to lose contact with the sidewalk? Another rock is
sitting on the concrete bottom of a swimming pool full of water.
The earthquake produces only vertical motion, so
the water does not slosh from side to side. (b) Present a
convincing argument that when the ground vibrates with
the amplitude found in part (a), the submerged rock also
barely loses contact with the floor of the swimming pool.

Explanation / Answer

Assuming constant amplitude (which is a good enough approximation for what I will do IF amplitude changes gradually). x(t) = A sin (omega t), so max x = A v(t) = A omega cos (omegat) so max v = A omega a(t) = -A omega^2 sin (omega t), so max a = A omega^2 The rock lets go when A omega^2 > g A > g / omega^2 = g / (2 pi f)^2 They give you f. You know g. Plugnchug. The rock in the pool has a different effective g. The net weight of the rock is: mg - (rhowater)gV = rhorock gV = (rhowater)gV , so the acceleration is g - rhowater gV/m = g (1- rhowater / rhorock). Use that instead of just g in the problem above. It is smaller, so it will take less amplitude to get the rock in the pool off the bottom than it took to get the other rock off the ground.

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