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gerryliyana

  • 2 years 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

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  1. gerryliyana
    • 2 years ago
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    @mukushla

  2. wilson3
    • 2 years ago
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    slow down buddy . you wrote u=f(x-vt) twice

  3. gerryliyana
    • 2 years ago
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    i wrote once, u = f(x - vt) and u = f(x + vt)

  4. Meepi
    • 2 years ago
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    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
    • 2 years ago
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    good job @Meepi i appreciate

  6. gerryliyana
    • 2 years ago
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    wait a sec

  7. gerryliyana
    • 2 years ago
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    isn't it 1/v2 ??

  8. Meepi
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    i thought \(\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\)

  11. gerryliyana
    • 2 years ago
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    ohh my bad.., sorry

  12. gerryliyana
    • 2 years ago
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    \[\frac{ ∂^{2}u }{ ∂x^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂t^{2} }\]

  13. Meepi
    • 2 years ago
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    It's alright your text book probably has the partial derivatives swapped

  14. Meepi
    • 2 years ago
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    yeah

  15. gerryliyana
    • 2 years ago
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    ur allright :), cool thanks :)

  16. Meepi
    • 2 years ago
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    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
    • 2 years ago
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    Same can be done for u = f(x - vt)

  18. gerryliyana
    • 2 years ago
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    ok.., i got it now :)

  19. Meepi
    • 2 years ago
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    awesome :D

  20. gerryliyana
    • 2 years ago
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    wanna help me with Legendre series ??

  21. Meepi
    • 2 years ago
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    Haven't really done much of that but I can give it a shot :)

  22. gerryliyana
    • 2 years ago
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    go to http://openstudy.com/study#/updates/5172d293e4b0f872395bc5be

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