Determine the turning points of the movement and the period of the oscillation of the pwrite-up.
I have actually began learning for my exam and also this is among the exercises in my textbook. Never before faced this type of question, so these are my thoughts so far:
To get the turning points I was reasoning of resolving the equation $E=V_0(e^-2ax-2e^-ax)$. I was doing the arithmetics with the absolute worth of $E$. But still I couldn"t seem to uncover the values for $x$. At the end I used Wolfram Alpha to uncover the worths however it provided me results via facility values. Is tright here a basic way to resolve this type of equations for $x$?
Anymethod, around the duration, I assume it"s the moment it takes for the pshort article to acquire from $x_1$ to $x_2$. But just how am I meant to method this? How would certainly I gain a time worth simply out of the equation for the potential?
I hope someone deserve to assist me out right here.
homework-and-exercises classical-mechanics potential-energy
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edited Jun 4 "15 at 9:47
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asked Jun 4 "15 at 8:18
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Energy conservation dictates$$ E = frac12mdotx^2 + V(x) = extconst$$With some arithmetic it follows$$ dotx = fracdxdt = sqrt2m^-1(E-V(x))$$This ODE can be addressed through separation of variables,yielding$$ int_t_1^t_2dt = int_x_1^x_2 fracdxsqrt2m^-1(E-V(x))$$The integral on the left hand also side can be evaluated automatically, wbelow $t_1$ and also $t_2$ are construed as the times as soon as the ppost is at $x_1$ or $x_2$ respectively. So it is simply fifty percent the duration.
Observe that the integral on the best diverges once $x$ ideologies the turning allude $E=V(x)$.
This strategy of addressing Newloads equations in a 1d potential have to be treated in any textbook on mechanics.
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For a basic potential it is in general hard or impossible to discover the turning points in closed form. Here but, you have the right to substitute $y=exp(-ax)$ and also resolve the corresponding quadratic equation. I"ll leave the explicit calculation to you.
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edited Jun 4 "15 at 10:54
answered Jun 4 "15 at 9:34
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nephente"s answer solves for the period. My answer is simply made to make you watch exactly how to go on from the suggest you were, solving:
$$ e^-2ax-2e^-ax=fracEV_0 $$
We make the adjust $y=e^-ax$, which yields:
$$y^2-2y-fracEV_0 = 0$$
2nd grade equation. Solve for $y$, and have actually in mind that $E/V_0$ is a negative number. It may provide you facility services otherwise.
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answered Jun 4 "15 at 11:01
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