## 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

1. gerryliyana

@mukushla

2. wilson3

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

3. gerryliyana

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

4. Meepi

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

good job @Meepi i appreciate

6. gerryliyana

wait a sec

7. gerryliyana

isn't it 1/v2 ??

8. Meepi

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

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

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

11. gerryliyana

12. gerryliyana

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

13. Meepi

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

14. Meepi

yeah

15. gerryliyana

ur allright :), cool thanks :)

16. Meepi

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

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

18. gerryliyana

ok.., i got it now :)

19. Meepi

awesome :D

20. gerryliyana

wanna help me with Legendre series ??

21. Meepi

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

22. gerryliyana