- #1
stunner5000pt
- 1,455
- 2
Of the partial kind
We have just finished solving the one dimensional wave equation for a vibrating string of length L using d'Alembert's solution which is
[tex] u(x,t) = \frac{1}{2} \displaystyle{[} f(x+ct) + f(x-ct) \displaystyle{]} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds [/tex]
the question is
If g(x) satisifes the conditions taht g(x) = -g(-x) and g(x) = -g(2l -x) then
[tex] \int_{x}^{x+2l} g(s) ds = 0 [/tex]
Well according to a theorem when a periodic function is integrated over a symmetric interval is zero. Also since the velocity of the wave is ct then ct = 2L
[tex] \int_{x-ct}^{x+ct} g(s) ds = 0 [/tex]
[tex] \int_{x-ct}^{x} g(s) ds + \int_{x}^{x+ct} g(s) ds = 0 [/tex]
[tex] \int_{x-2L}^{x} g(s) ds = -\int_{x}^{x+2L} g(s) ds [/tex]
but here is where i am stuck. How do i change the limit of integration over the left hand side
ALso I am not even sure is if this is the correct way to go? Is it the right way? Your help and advice would be greatly appreciated!
Thank you!
We have just finished solving the one dimensional wave equation for a vibrating string of length L using d'Alembert's solution which is
[tex] u(x,t) = \frac{1}{2} \displaystyle{[} f(x+ct) + f(x-ct) \displaystyle{]} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds [/tex]
the question is
If g(x) satisifes the conditions taht g(x) = -g(-x) and g(x) = -g(2l -x) then
[tex] \int_{x}^{x+2l} g(s) ds = 0 [/tex]
Well according to a theorem when a periodic function is integrated over a symmetric interval is zero. Also since the velocity of the wave is ct then ct = 2L
[tex] \int_{x-ct}^{x+ct} g(s) ds = 0 [/tex]
[tex] \int_{x-ct}^{x} g(s) ds + \int_{x}^{x+ct} g(s) ds = 0 [/tex]
[tex] \int_{x-2L}^{x} g(s) ds = -\int_{x}^{x+2L} g(s) ds [/tex]
but here is where i am stuck. How do i change the limit of integration over the left hand side
ALso I am not even sure is if this is the correct way to go? Is it the right way? Your help and advice would be greatly appreciated!
Thank you!