Conservation of power in a traveling wave on a string

In summary: The solution should have mentioned that ##x = 0## is taken to be where the two strings are attached. Otherwise, the boundary condition mentioned above would not hold at all times if you use the expressions for ##y_I##, ##y_R##, and ##y_T## at the bottom of page 1 of the...In summary, the textbook's solution is incorrect. There are several errors, and the solution does not make sense.
  • #1
Adrian Simons
10
4
Homework Statement
This is Problem #71 in Chapter 15 of Paul A. Tipler and Gene Mosca, PHYSICS For Scientists and Engineers, Sixth Edition,
W. H. Freeman & Co., New York, NY, 2008.
Relevant Equations
$$1 = r^2 + \left( \frac{v_1}{v_2} \right) \tau^2$$ where ##\tau## and ##r## are the transmission and reflection coefficients given by $$\tau = \frac{2 v_2}{v_2 + v_1}$$ and $$r = \frac{v_2 - v_1}{v_2 + v_1}$$.
The statement of the problem is:

Consider a taut string that has a mass per unit length ##\mu_1## carrying transverse wave pulses of the form ##y = f(x - v_1 t)## that are incident upon a point P where the string connects to a second string with mass per unit length ##\mu_2##.
Derive $$1 = r^2 + \left( \frac{v_1}{v_2} \right) \tau^2$$ by equating the power incident on point P to the power reflected at P plus the power transmitted at P.

The solution given in the solutions manual to the textbook is wrong. There is one glaring error in it, in addition to what I believe are some more subtle errors. Also, there are several things they do without any motivation for why they're doing it, which I believe are incorrect. Otherwise, I've been unable to solve the problem. Can anyone provide a viable solution?
 
Last edited:
Physics news on Phys.org
  • #2
Please post the book solution and point out where you believe the first error is.
 
  • Like
Likes MatinSAR
  • #3
Sorry, I cannot decode the macros.
Can you try posting it directly as LaTeX?
 
  • Like
Likes MatinSAR
  • #4
I'm so sorry. I'm an expert in LaTex, but I'm a novice when it comes to MathJax, and I'm having a lot of trouble posting it. What I'm going to do is to write it up in Latex and post it as an attachment. Please be patient.
 
  • #5
haruspex said:
Please post the book solution and point out where you believe the first error is.
Please open the attached file in which I have outlined the solution given in the Instructor's Solutions Manual. I have also noted in the document where I believe the errors occur.
 

Attachments

  • Ch15_P71.pdf
    61.1 KB · Views: 78
  • #6
Adrian Simons said:
Please open the attached file in which I have outlined the solution given in the Instructor's Solutions Manual. I have also noted in the document where I believe the errors occur.
Could the author be using a convention where the sign (+/-) of the power’s value indicates the direction of energy flow?

Although this is inconsistent with the wording in the problem statement, it would make sense of ##P_I + P_R = P_T## since the value of ##P_R## would be negative. Then the offending part of the problem statement is equivalent to ##|P_I| = |P_R| + |P_T|##.
 
  • Like
Likes MatinSAR
  • #7
If you look at the rest of the solution, I don't see where they treat things the way you imply they're doing. Yet, they magically come up with the correct formula in the end. And even even if what you say were true, that doesn't do anything to explain the rest of the problems with their solution. But thank you for your response.
 
  • #8
Adrian Simons said:
If you look at the rest of the solution, I don't see where they treat things the way you imply they're doing. Yet, they magically come up with the correct formula in the end. And even even if what you say were true, that doesn't do anything to explain the rest of the problems with their solution. But thank you for your response.
I believe the power transmitted is given by ##P = -T\frac {∂y}{∂x} \frac {∂y}{∂t}## irrespective of the wave direction. If ##P>0##, power flows in the +x direction; if ##P<0##, power flows in the -x direction.

Illustration…

Consider #he wave ##y = \cos(x - vt)## which propagates in the +x direction.
##P = -T\frac {∂y}{∂x} \frac {∂y}{∂t} = -T (-\sin(x-vt))~(-v(-\sin(x - vt))) = vT\sin^2(x-vt)##
This is positive, indicating power flows in the +x direction.

Now consider the wave ##y = \cos(x + vt)## which propagates in the -x direction.
##P = -T\frac {∂y}{∂x} \frac {∂y}{∂t} = -T (-sin(x + vt))~(v(-sin(x + vt))) = -vTsin^2(x + vt)##
This is negative indicating power flows in the -x direction.

Section '3 Energy Flux' here may be useful: https://users.physics.ox.ac.uk/~palmerc/Wavesfiles/Energy_Handout.pdf

Edited to fix mismatched brackets.
 
  • Like
Likes MatinSAR
  • #9
##y## for string 1 and ##y## for string 2 must have the same time dependence at the point where they are joined together. Otherwise, they couldn't remain joined together as time passes. The factor ##\large \frac {v_1}{v_2}## in the argument of ##f## for the transmitted wave is necessary to satisfy this condition. The time term in the argument of ##f## for all three waves is ##-v_1 t##. The solution that you quoted should have explained this if the textbook doesn't discuss it.

The factor of ##\large \frac {v_1}{v_2}## for the transmitted wave means that the shape of the wave that is transmitted is not the same as the shape of the incident wave. For example, if the incoming wave is sinusoidal with wavelength ##\lambda_1##, the transmitted wave will be sinusoidal with wavelength ##\lambda_2 = \frac {v_2}{v_1} \lambda_1##.

I think the solution should have mentioned that ##x = 0## is taken to be where the two strings are attached. Otherwise, the boundary condition mentioned above would not hold at all times if you use the expressions for ##y_I##, ##y_R##, and ##y_T## at the bottom of page 1 of the solution.

The factors ##\frac {\partial f}{\partial \eta_I}##, ##\frac {\partial f}{\partial \eta_R}##, and ##\frac {\partial f}{\partial \eta_T}## that are canceled near the end of the solution are all equal to one another when evaluated at the junction point ##x = 0##. At ##x = 0##, ##\eta_I = \eta_R = \eta_T = -v_1t##.
 
  • Like
Likes MatinSAR
  • #10
TSny said:
##y## for string 1 and ##y## for string 2 must have the same time dependence at the point where they are joined together. Otherwise, they couldn't remain joined together as time passes. The factor ##\large \frac {v_1}{v_2}## in the argument of ##f## for the transmitted wave is necessary to satisfy this condition. The time term in the argument of ##f## for all three waves is ##-v_1 t##. The solution that you quoted should have explained this if the textbook doesn't discuss it.

The factor of ##\large \frac {v_1}{v_2}## for the transmitted wave means that the shape of the wave that is transmitted is not the same as the shape of the incident wave. For example, if the incoming wave is sinusoidal with wavelength ##\lambda_1##, the transmitted wave will be sinusoidal with wavelength ##\lambda_2 = \frac {v_2}{v_1} \lambda_1##.

I think the solution should have mentioned that ##x = 0## is taken to be where the two strings are attached. Otherwise, the boundary condition mentioned above would not hold at all times if you use the expressions for ##y_I##, ##y_R##, and ##y_T## at the bottom of page 1 of the solution.

The factors ##\frac {\partial f}{\partial \eta_I}##, ##\frac {\partial f}{\partial \eta_R}##, and ##\frac {\partial f}{\partial \eta_T}## that are canceled near the end of the solution are all equal to one another when evaluated at the junction point ##x = 0##. At ##x = 0##, ##\eta_I = \eta_R = \eta_T = -v_1t##.
The solution discussed none of the things you mention, but this is me for not realizing these things for myself. Thank you so much for your help.
kicking-myself-image-man-himself-37498981.jpg
 
  • Haha
  • Like
Likes TSny and berkeman

What is conservation of power in a traveling wave on a string?

Conservation of power in a traveling wave on a string refers to the principle that the total power of a wave remains constant as it travels along a string. This means that the energy of the wave is conserved and does not increase or decrease as it propagates.

How is power calculated in a traveling wave on a string?

The power of a traveling wave on a string is calculated by multiplying the amplitude of the wave by the velocity of the wave. This is known as the instantaneous power, as it represents the amount of energy being transferred at a specific point in time.

What factors affect the conservation of power in a traveling wave on a string?

The conservation of power in a traveling wave on a string is affected by the tension of the string, the amplitude of the wave, and the velocity of the wave. A higher tension or amplitude can result in a higher power, while a slower wave velocity can decrease the power.

Why is conservation of power important in a traveling wave on a string?

Conservation of power is important in a traveling wave on a string because it allows us to understand and predict the behavior of the wave. It also helps us to conserve energy and ensure that the wave does not lose or gain power as it travels.

How does conservation of power apply to real-life situations?

Conservation of power in a traveling wave on a string has many real-life applications, such as in musical instruments like guitars and pianos, where the strings vibrate to produce sound waves. It also applies to telecommunications, where signals are transmitted through cables or fibers as traveling waves. In both cases, the conservation of power ensures that the signals remain strong and do not lose power as they travel.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
854
  • Introductory Physics Homework Help
Replies
2
Views
805
Replies
1
Views
961
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
838
Back
Top