How Does Hooke's Law Affect Wave Propagation in Stretched Rubber Bands?

[DivGradCurl]

Hi,

Could you please give a hand with this question from the book "Fundamentals of Physics/Halliday, Resnick, Walker" - 6th ed, page 395, #22P. Here it goes:

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The type of rubber band used inside some baseballs and golf balls obeys Hooke's Law over a wide range of elongation of the band. A segment of this material has an ustretched length l and mass m. When a force F is applied, the band stretches an additional length Δl.

(a) What is the speed (in terms of m, Δl, and spring constant k) of transverse waves on this stretched rubber band?

(b) Using your answer to (a), show that the time required for a transverse pulse to travel the length of the rubber band is proportional to 1/&radic;(&Delta;l) if &Delta;l << l and constant if &Delta;l >> l

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Comments:

(a) v = &radic;([tau]/[mu]), where [tau] = F = k&Delta;l and [mu] = m/(l+&Delta;l). This gives: v = &radic;{[k&Delta;l(l+&Delta;l)]/m} --- This should be right.

(b) I'm not sure what happens in either case... my guess is that when:

&Delta;l << l, we have: v = &radic;[(kl)/m].

v = dl/dt Then: [&int;(0,&Delta;l)] dt = [&int;(0,&Delta;l)] (1/v) dl

This gives: t = [2&radic;(&Delta;l)]/&radic;(k;m)

That doesn't seem to fit, but could be close.

&Delta;l >> l, we have: v = &radic;{[k(&Delta;l)^2]/m}.

v = dl/dt Then: [&int;(0,&Delta;l)] dt = [&int;(0,&Delta;l)] (1/v) dl

This gives: t = [1/&radic;(k;m)]

That doesn't seem to fit, but could be close as well.

Thanks a lot!

 

[DivGradCurl]

Just a correction

Over comment for (b) [squ](k;m) = [squ](k/m)

 

[M98Ranger]

Your approach to part (a) is correct. The speed of the transverse wave is indeed given by v = &radic;([k&Delta;l(l+&Delta;l)]/m).

For part (b), you are on the right track. The main concept to understand here is that the speed of the wave is directly proportional to the square root of the tension (k) and inversely proportional to the square root of the mass per unit length ([mu] = m/(l+&Delta;l)).

In the case where &Delta;l << l, we can assume that l+&Delta;l is approximately equal to l. Therefore, the mass per unit length becomes [mu] = m/l. Plugging this into the equation for v, we get: v = &radic;([k&Delta;l(l+&Delta;l)]/m) = &radic;([k&Delta;l(l)]/m) = &radic;(kl/m). This is the same as your answer for the speed in this case.

Now, to find the time for the pulse to travel the length of the rubber band, we use the equation v = dl/dt. Rearranging for dt, we get: dt = dl/v. Integrating this from 0 to &Delta;l, we get: t = [&int;(0,&Delta;l)] dl/v = [&int;(0,&Delta;l)] dl/&radic;(kl/m). Simplifying this, we get: t = [2&radic;(&Delta;l)]/&radic;(k;m). This matches your answer and shows that the time is proportional to the square root of &Delta;l in this case.

In the case where &Delta;l >> l, we can assume that &Delta;l is much larger than l. Therefore, the mass per unit length becomes [mu] = m/&Delta;l. Plugging this into the equation for v, we get: v = &radic;([k&Delta;l(l+&Delta;l)]/m) = &radic;([k&Delta;l(&Delta;l)]/m) = &radic;[k(&Delta;l)^2/m]. This matches your answer for the speed in this case.

Following the same steps as before, we get: t = [&int;(0,&Delta;l)] dl/v = [&int;(0,&Delta;l)] dl/&rad