Tag Info

New answers tagged

1

You can show this by noting that $k_1^2=k_2^2$ and $X(x)T(t)= F(k_1x\pm k_2vt)$. You can see that by: $$ { \partial^2 f \over \partial t^2 } = v^2 { \partial^2 f \over \partial x^2 } $$ Is more common take $k_1=k$ and $k_2v=-\omega$. Then you may note that the equation imposes the choices of $k_1$ and $k_2$, or $k$ and $\omega$. The imposes is ...


1

Hint: Use light cone coordinates. What is the full solution to $$ \frac{\partial^2 f(x^{+},x^{-})}{\partial x^{+}\partial x^{-}}~=~0~?$$


0

Consider showing that $f_L$ and $f_R$ are themselves solutions no matter what function (of one variable) they are. Next consider any particular solution (i.e. with particular boundary conditions). Then try relating your boundary conditions to a particular $f_L$ and $f_R$ jointly. If you have a result about uniqueness, then you know exactly how many ...


2

You can treat the friction term as a perturbation. During all the stages of the motion, the effect of the friction is always small on small time scales. So, it always looks like as if the motion exactly fits the solution of the differential equation where b is set to zero. However, over long periods of time the value of the integration constants will slowly ...


0

UPDATE: While I was typing the answer below, I see that you came to the same conclusion as I do in my answer below. From your $\dot \theta$ equation, it appears that a positive $x$ component of force acts to increase the rate of change of the angle when $0 \lt \theta \lt \pi$. But, that can't be correct. If you set $g$ and $b$ to zero and launch the ...


0

Every differential equation problem has two parts: the differential equation itself, and the boundary condition. Neither can tell you the other in and of itself. Consider $y'' = -k^2 y$. Is the solution sine or cosine? The answer depends on boundary conditions, and I have to specify them. I can't get them by specifying anything in the equation. So, say your ...



Top 50 recent answers are included