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Title: GR8677.036
Post by: conantee on January 02, 2010, 06:56:30 AM
Question 34-36
The potential energy of a body constrained to move on a straight line is kx^4 where k is a constant. The position of the body is x, its speed v, its linear momentum p, and its mass m.

36. The body moves from x_1 at time t_1 to x_2 at time t_2. Which of the following quantities is an extremum for the x-t curve corresponding to this motion, if end points are fixed?
(A) \int_{t_1}^{t_2} (\frac{1}{2}mv^2 - kx^4) dt
(B) \int_{t_1}^{t_2} (\frac{1}{2}mv^2) dt
(C) \int_{t_1}^{t_2} (mxv) dt
(D) \int_{t_1}^{t_2} (\frac{1}{2}mv^2 + kx^4) dt
(E) \int_{x_1}^{x_2} (mv) dx


Title: Re: GR8677.036
Post by: conantee on January 02, 2010, 07:11:10 AM
If you can remember what action is, you've done. If not, you are in trouble. However, you can cross (B) out since it is too plain (just kinetic energy?).

The correct answer is (A). The time integration of Lagrangian (L = T-V) , called action, is extremum (This theory is called principle of least action: particles choose the path such that the action is extremum).

41 of 100 people got correct answer.