- #1

LCSphysicist

- 634

- 153

- Homework Statement:
- I don't know why, but my answer is wrong.

- Relevant Equations:
- .

"Consider a string of length L that is connected at both ends to supports and is subjected to a load (external force per unit length) of f(x). Find the displcament u"

https://i.stack.imgur.com/yVIDG.png

We need to solve this:

$$Tu_{xx} = f(x)$$ subject to $$u(0)=u(L)=0$$

But i don't understand what is the problem in my solutions:

$$Tu_{xx} = f(x) = > Tu_{xx} = \delta({x-\epsilon})\\T(Us^2-su(0)-u_{x}(0)) = e^{-s \epsilon}\\U = (e^{-s \epsilon}/T + u_{x}(0))/s^2$$

So, we know that the inverse of $$e^{-s \epsilon}/(Ts^2) = H(x-\epsilon)x/T$$ and inverse of $$(u_{x}(0))/s^2 = u_{x}(0)x$$

So, shouldn't the final answer be $$u = \int_{0}^{L} f(\epsilon) (H(x-\epsilon)x/T + u_{x}(0)x) d \epsilon$$

Why is it wrong?

https://i.stack.imgur.com/yVIDG.png

We need to solve this:

$$Tu_{xx} = f(x)$$ subject to $$u(0)=u(L)=0$$

But i don't understand what is the problem in my solutions:

$$Tu_{xx} = f(x) = > Tu_{xx} = \delta({x-\epsilon})\\T(Us^2-su(0)-u_{x}(0)) = e^{-s \epsilon}\\U = (e^{-s \epsilon}/T + u_{x}(0))/s^2$$

So, we know that the inverse of $$e^{-s \epsilon}/(Ts^2) = H(x-\epsilon)x/T$$ and inverse of $$(u_{x}(0))/s^2 = u_{x}(0)x$$

So, shouldn't the final answer be $$u = \int_{0}^{L} f(\epsilon) (H(x-\epsilon)x/T + u_{x}(0)x) d \epsilon$$

Why is it wrong?