Stress, Strain, and Sound in a Projectile Steel Rod

[Ghost Repeater]

Homework Statement

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For a certain type of steel, stress is always proportional to strain with Young's modulus 20 x 10^10 N/m^2. The steel has density 7.86 x 10^3 kg/m^3. A rod 80.0 cm long, made of this steel, is fired at 12.0 m/s straight at a very hard wall.

a) The speed of a one-dimensional compressional wave moving along the rod is given by v = sqrt(Y/rho), where Y is Young's modulus for the rod and rho is the density of steel. Calculate this speed.

b) After the front end of the rod hits the wall and stops, the back end of the rod keeps moving as described by Newton's first law until it is stopped by excess pressure in a sound wave moving back through the rod. What time interval elapses before the back end of the rod receives the message that it should stop?

Homework Equations

The Attempt at a Solution

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This is a problem from Serway & Jewett's Physics for Scientists and Engineers textbook. Chapter 17, Problem 59.

I know the solution to the problem (from the solutions manual) but I have a question about the solution, because it doesn't make sense to me.

Part a) is straightforward of course. Just plug in the given data to the formula v = sqrt(Y/rho). But I am perplexed by part b).

The solution given to part b) by the authors is this: "The signal to stop passes between layers of atoms as a sound wave, reaching the back end of the bar in time interval Δt = L/v. As described by Newton's first law, the rearmost layer of steel has continued to move forward with its original speed v for this time, compressing the bar by ΔL = v_i *Δt."

This is the part I do not understand. To calculate the compression ΔL, the authors use the time it takes the wave to pass the entire uncompressed length of the bar. But isn't the whole point of the compression that the back end of the bar is moving toward the wave, at the same time as the wave moves toward the back end of the bar? So wouldn't it take the wave less time to meet the back end of the bar, since the two are approaching each other?

The way I tried to solve the problem was to treat the wave as moving from the front to the rear of the bar, and the back end of the bar as moving from the back end to the front, and the two would meet somewhere in the middle. But if the bar is compressing, then the wave can't take the whole time it would take to travel the uncompressed length to reach the compressed length, right?

Thanks for helping me clear up confusion.

 

[trurle]

This is the part I do not understand. To calculate the compression ΔL, the authors use the time it takes the wave to pass the entire uncompressed length of the bar. But isn't the whole point of the compression that the back end of the bar is moving toward the wave, at the same time as the wave moves toward the back end of the bar? So wouldn't it take the wave less time to meet the back end of the bar, since the two are approaching each other?

This effect would make a very small correction (about 0.2% at 12 m/s projectile speed). I think you are expected to neglect it.

 

[Ghost Repeater]

Well, that makes sense! Negligible is negligible, after all, ha ha. Thanks for the reply.