Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

gerryliyana

  • one year ago

show that the expression u = sin (x-vt) describing a sinusoidal wave, satisfies the wave equation. Show that in general u = f(x - vt) and u = f(x + vt) satisfies the wave equations, where f is any function with a second derivative

  • This Question is Closed
  1. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @mukushla

  2. wilson3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    slow down buddy . you wrote u=f(x-vt) twice

  3. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i wrote once, u = f(x - vt) and u = f(x + vt)

  4. Meepi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The wave equation is: \[\frac{\partial^2 u}{\partial t^2} = v^2\frac{\partial^2 u}{\partial x^2}\] In your case, \(u = \sin(x - vt)\) And \[u_{tt} = -\sin(x - vt) * v^2 \] \[u_{xx} = -\sin(x - vt)\] So since \(u_{tt} = v^2 u_{xx}\), u satifies the wave equation

  5. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    good job @Meepi i appreciate

  6. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    wait a sec

  7. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    isn't it 1/v2 ??

  8. Meepi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    No, \(u_{tt} = -\sin(x - vt) * v^2\) and \(u_{xx} = -\sin(x - vt)\) the v^2 is from the chain rule when you derive with respect to t in \(u_{tt}\) since \(u_{xx} = -\sin(x - vt)\), \(u_{tt} = u_{xx} * v^2\)

  9. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    isn't it \[\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\] ??? you typed \(\frac{ ∂^{2}u }{ ∂t^{2} } = v^{2} \frac{ ∂^{2}u }{ ∂x^{2} } \)

  10. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i thought \(\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\)

  11. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ohh my bad.., sorry

  12. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac{ ∂^{2}u }{ ∂x^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂t^{2} }\]

  13. Meepi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    It's alright your text book probably has the partial derivatives swapped

  14. Meepi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah

  15. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ur allright :), cool thanks :)

  16. Meepi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Also, for the general case, just use the chain rule as well to get \[u_{xx} = f''(x + vt)\] \[u_{tt} = f''(x + vt) * v^2\] Then \[u_{xx} = \frac{u_{tt}}{v^2}\]

  17. Meepi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Same can be done for u = f(x - vt)

  18. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok.., i got it now :)

  19. Meepi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    awesome :D

  20. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    wanna help me with Legendre series ??

  21. Meepi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Haven't really done much of that but I can give it a shot :)

  22. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    go to http://openstudy.com/study#/updates/5172d293e4b0f872395bc5be

  23. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.