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This question is about mechanical waves in solids, the speed at which forces propagate through solids, and the displacement that might occur due to unequal starting times of equally sized forces.

There is a metal box which contains two signal paths: a short one and a long one. Forces that can be generated (for example by a hammer) can travel through these paths. See the initial state in the image. Left is the long signal path, and on the right there's the short signal path. The waves are travelling some distance marked by the dotted yellow and light green lines on the blue box's edges.

metal box with different time signal paths

The red arrows represent forces. The plates next to it are the spots where the hammers hit.

In situation a), the forces are equal in size and opposite to each other. Force 1 starts earlier than force 2. A few milliseconds later, force 2 starts. It's just a short hammer blast on both sides, just a smash and release. The left side of the box is now being pulled to the left, and the right side is pulled to the right. After some time, the mechanical waves meet in the middle. The forces travel at a finite speed; the speed of sound in the material.

In situation b), the waves don't meet in the center but near the left edge. The times that the forces start are still unequal. Force 1 still starts before force 2, but now with even less time in between.

My questions are:

  • Even though the forces started at different times, is there any displacement of the metal box in any of the situations? Or is there any movement at all but is the net displacement zero?
  • What happens when the waves (signals) meet? Do they annihilate and communicate to all particles that the velocity must become zero throughout the whole system?
  • How can I visualize the movement of the particles of the box? For example in situation a), is the left side moving to the left, while the right side moves to the right and the center staying fixed? Is this like a stretching effect of the box?
  • Does the left side of the box (with the longer signal path) affect the right side of the box by pulling harder to the left than the right side does because it is more massive, and thus creating a displacement to the left?

Thanks for reading.

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Even though the forces started at different times, is there any displacement of the metal box in any of the situations? Or is there any movement at all but is the net displacement zero?

Sure. If you think of each force as causing an acceleration, the first one begins an acceleration in one direction, the second an acceleration in the other (or a deceleration). The total displacement depends on the time difference between them.

What happens when the waves (signals) meet? Do they annihilate and communicate to all particles that the velocity must become zero throughout the whole system?

Waves pass through each other and will here as well. I would expect the entire system to vibrate and "ring". Over time, the waves will dissipate as heat and the vibrations will be attenuated. If we think of the strikes as a particular impulse ($\Delta p$) and they are equal and opposite, then the total momentum of the box must remain zero and the box must end up stationary.

How can I visualize the movement of the particles of the box? For example in situation a), is the left side moving to the left, while the right side moves to the right and the center staying fixed? Is this like a stretching effect of the box?

Yes. Think of the components as stiff springs. They will vibrate, stretch, and compress in response to forces. All materials must do this.

Does the mass of the left side (longer signal path) affect the right side by pulling harder and thus creating a displacement to the left?

Pulling harder than what?

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  • $\begingroup$ Thank you! But without knowing the values of forces in situation a), we only know that F1 starts before F2. Because the signals meet in the middle, the box stretches to both sides. When the waves meet, will the stretching stop and will the box fall back into its original shape? And can you tell if there is any displacement to the left or right? Or no displacement due to both sides stretching equally? $\endgroup$ – Arundel Aug 20 '14 at 13:42
  • $\begingroup$ My last question meant to be: does the left side of the box (with the longer signal path) affect the right side of the box by pulling harder to the left than the right side does because it is more massive, and thus creating a displacement to the left? $\endgroup$ – Arundel Aug 20 '14 at 13:45
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    $\begingroup$ Unless there is permanent (plastic) deformation, the box will return to the original shape. The timing has no effect on that. The displacement depends on the timing. Once one side is given an impulse, the center of mass must begin moving. When struck in the opposite direction (per your specifications), the center of mass must stop moving. That gives you a displacement. $\endgroup$ – BowlOfRed Aug 20 '14 at 15:53
  • $\begingroup$ Both sides are pulling on each other. This is a pair of action/reaction forces and must therefore be equal and opposite. Neither side can pull "harder". $\endgroup$ – BowlOfRed Aug 20 '14 at 15:54
  • $\begingroup$ Ok thank you for clarifying my comment and my last question that I rephrased. Ok you say that there will be a displacement depending upon timing, and that neither side can pull "harder". Is there a situation where there will be no net displacement, even though the times are different? Also, does the size of these equal but opposite forces have to do with the amount of displacement? $\endgroup$ – Arundel Aug 20 '14 at 18:27

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