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 one year ago
show that the expression u = sin (xvt) 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
 one year ago
show that the expression u = sin (xvt) 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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wilson3
 one year ago
Best ResponseYou've already chosen the best response.0slow down buddy . you wrote u=f(xvt) twice

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1i wrote once, u = f(x  vt) and u = f(x + vt)

Meepi
 one year ago
Best ResponseYou've already chosen the best response.1The 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

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1good job @Meepi i appreciate

Meepi
 one year ago
Best ResponseYou've already chosen the best response.1No, \(u_{tt} = \sin(x  vt) * v^2\) 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 \(u_{xx} = \sin(x  vt)\), \(u_{tt} = u_{xx} * v^2\)

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1isn't it \[\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\] ??? you typed \(\frac{ ∂^{2}u }{ ∂t^{2} } = v^{2} \frac{ ∂^{2}u }{ ∂x^{2} } \)

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1i thought \(\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\)

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1ohh my bad.., sorry

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ ∂^{2}u }{ ∂x^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂t^{2} }\]

Meepi
 one year ago
Best ResponseYou've already chosen the best response.1It's alright your text book probably has the partial derivatives swapped

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1ur allright :), cool thanks :)

Meepi
 one year ago
Best ResponseYou've already chosen the best response.1Also, 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 \[u_{xx} = \frac{u_{tt}}{v^2}\]

Meepi
 one year ago
Best ResponseYou've already chosen the best response.1Same can be done for u = f(x  vt)

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1ok.., i got it now :)

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1wanna help me with Legendre series ??

Meepi
 one year ago
Best ResponseYou've already chosen the best response.1Haven't really done much of that but I can give it a shot :)

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1go to http://openstudy.com/study#/updates/5172d293e4b0f872395bc5be
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