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New answers tagged differential-equations

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One can consider the quantities $\int F_x\,dx=\int m\ddot{x}\,dx=\frac{1}{2}m(\dot{x}_f^2-\dot{x}_i^2)$ The $y$ version of above The $z$ version of above Are these what you're after? These three quantities aren't usually considered in standard problems, but they seem valid to me. Your "Result 2" is the sum of the three bulleted equations here.

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Consider the positive quantity $X = (\omega - \omega_0)^2 (\omega + \omega_0)^2$. Let us make the approximation $\omega \approx \omega_0$: In the first factor: we get $X_1 = 0$ and the relative error $\epsilon _X = |\frac{X_1-X}{X}| = 100 \%$. In the second factor: we get $X_2 = 4 \omega _0^2 (\omega - \omega_0)^2$ and $\epsilon _X = | \frac{X_2-X}{X}| = ... 0 If the particles are all the same (we don't need to invoke indistinguishability at that stage I think), then there is no loss of information since the permutation of two particles in a reduced distribution function will ask exactly the same question as before the permutation probability-wise. Note that usually, reduced distribution functions are defined ... 2 Here are the steps you can take. Degrees of Freedom. There are 3 degrees of freedom, one for the base plate, one for the box and one for the mass. Hence there are 3 variables that you need to track, as well as their derivatives. I will name them$x_0=\gamma(t)$for the plate,$x_1$for the box and$x_2$for the ball. Free Body Diagrams. For the moving ... 0 The behavior of the system (not surprisingly)depends on the initial conditions. (For the sake of argument, we can assume the box starts stationary with respect to the table and$\gamma(0)=0$) I am assuming the problem is$1d$; this way we will end up with two coupled equations of motion. Let's show the box's coordinate with$\chi\$ and ball's coordinate with ...

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