Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Differential Equations
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

slow down buddy . you wrote u=f(x-vt) twice
i wrote once, u = f(x - vt) and u = f(x + vt)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
good job @Meepi i appreciate
wait a sec
isn't it 1/v2 ??
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 \(u_{xx} = -\sin(x - vt)\), \(u_{tt} = u_{xx} * v^2\)
isn't it \[\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\] ??? you typed \(\frac{ ∂^{2}u }{ ∂t^{2} } = v^{2} \frac{ ∂^{2}u }{ ∂x^{2} } \)
i thought \(\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\)
ohh my bad.., sorry
\[\frac{ ∂^{2}u }{ ∂x^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂t^{2} }\]
It's alright your text book probably has the partial derivatives swapped
yeah
ur allright :), cool thanks :)
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 \[u_{xx} = \frac{u_{tt}}{v^2}\]
Same can be done for u = f(x - vt)
ok.., i got it now :)
awesome :D
wanna help me with Legendre series ??
Haven't really done much of that but I can give it a shot :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question