By A. Mous

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The period is just twice the time it takes to get between turning points. Partial answer: P eriod = π 2mωx2 + ω k0 11. The potential energy of a particle moving in one dimension is k2 1 V (x) = k1 x2 + , 2 x with k1 > 0, k2 > 0, and x > 0. What do you guess the frequency will be for small oscillations. Show that the motion is periodic. What is the equilibrium position? What is the frequency of the motion if the amplitude of the vibrations is very small? 12. , periodic motion) of a particle, the approximate differential equation of motion (obtained by doing a Taylor’s expansion of the potential function) is found to be (1 + a2 ) 4 3 d2 x + 3(x − x1 ) − (x − x1 )2 + 2 (x − x1 )3 = 0.

2. We may use the expression which we have just derived for the Christoffel symbols to write, ∂gin ∂gjn ∂gij 1 + − n glk Γkij = glk g kn 2 ∂q j ∂q i ∂q Observe, glk g kn = δln , 35 . so that we can write, 1 ∂gin ∂gjn ∂gij glk Γkij = δln + − n 2 ∂q j ∂q i ∂q = 1 2 ∂gil ∂gjl ∂gij + − ∂q j ∂q i ∂q l . 3. Thus, Al = glk q¨k + 1 2 ∂gil ∂gjl ∂gij + − ∂q j ∂q i ∂q l q˙i q˙j . In principle, the covariant components of the acceleration can be obtained from this expression once the covariant form of the metric tensor is known.

Rather, an equation of this type should be solved by actually carrying out the steps just illustrated, 45 1. Compute the integrating factor. 2. Multiply the differential equation by the integrating factor. 3. Integrate both sides of the equation, remembering that the left-handside is a perfect differential. 4. Solve the integrated equation for y(x). The first-order equation may occur in mechanics in the following way. If a frictional force proportional to the square of velocity exists, Newton’s Second Law may take the form, m¨ x = h(x) − g(x)x˙ 2 .