Question

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

Answer 1

@mukushla

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

good job @Meepi i appreciate

Answer 6

wait a sec

Answer 7

isn't it 1/v2 ??

Answer 8

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 \(u_{xx} = -\sin(x - vt)\), \(u_{tt} = u_{xx} * v^2\)

Answer 9

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

Answer 10

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

Answer 11

ohh my bad.., sorry

Answer 12

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

Answer 13

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

Answer 14

yeah

Answer 15

ur allright :), cool thanks :)

Answer 16

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 \[u_{xx} = \frac{u_{tt}}{v^2}\]

Answer 17

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

Answer 18

ok.., i got it now :)

Answer 19

awesome :D

Answer 20

wanna help me with Legendre series ??

Answer 21

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

Answer 22

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