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gerryliyana
Group Title
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
 one year ago
gerryliyana Group Title
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
 one year ago

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gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
@mukushla
 one year ago

wilson3 Group TitleBest ResponseYou've already chosen the best response.0
slow down buddy . you wrote u=f(xvt) twice
 one year ago

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

Meepi Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

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

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
wait a sec
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
isn't it 1/v2 ??
 one year ago

Meepi Group TitleBest ResponseYou've already chosen the best response.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 \(u_{xx} = \sin(x  vt)\), \(u_{tt} = u_{xx} * v^2\)
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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