The proof for ( x ) = ∫ ϕ * ( i h ∂ ∂ p ) ϕ d p .

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A pendulum of length l and mass m is mounted on a block of mass M. The block can movefreely without friction on a horizontal surface as shown in Fig 1 PHYS 24021. Consider a particle of mass m moving in a plane under the attractive force μm/r2 directedto the origin of polar coordinates r, θ. Determine the equations of motion.2. Write down the Lagrangian for a simple pendulum constrained to move in a single verticalplane. Find from it the equation of motion and show that for small displacements fromequilibrium the pendulum performs simple harmonic motion.3. A pendulum of length l and mass m is mounted on a block of mass M. The block can movefreely without friction on a horizontal surface as shown in Fig 1.Figure 1Show that the Lagrangian for the system isL =( M + m2)( ̇x)2 + ml ̇x ̇θ + m2 l2( ̇θ)2 + mgl(1 − θ22)

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Chapter 3.4, Problem 3.13P

To determine

The proof for (x)=∫ϕ*(ih∂∂p)ϕdp.

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A pendulum of length l and mass m is mounted on a block of mass M. The block can movefreely without friction on a horizontal surface as shown in Fig 1 PHYS 24021. Consider a particle of mass m moving in a plane under the attractive force μm/r2 directedto the origin of polar coordinates r, θ. Determine the equations of motion.2. Write down the Lagrangian for a simple pendulum constrained to move in a single verticalplane. Find from it the equation of motion and show that for small displacements fromequilibrium the pendulum performs simple harmonic motion.3. A pendulum of length l and mass m is mounted on a block of mass M. The block can movefreely without friction on a horizontal surface as shown in Fig 1.Figure 1Show that the Lagrangian for the system isL =( M + m2)( ̇x)2 + ml ̇x ̇θ + m2 l2( ̇θ)2 + mgl(1 − θ22)

Divergence theorem. (a) Use the divergence theorem to prove, v = -478 (7) (2.1) (b) [Problem 1.64, Griffiths] In case you're not persuaded with (a), try replacing r by (r² + e²)2 and watch what happens when ɛ → 0. Specifically, let 1 -V². 4л 1 D(r, ɛ) (2.2) p2 + g2 By taking note of the defining conditions of 8°(7) [(1) at r = 0, its value goes to infinity, (2) for all r + 0, its value is 0, and (3) the integral over all space is 1], demonstrate that 2.2 goes to 8*(F) as ɛ → 0.

2.7 There are certain simple one-dimensional problems where the equation of motion (Newton's second law) can always be solved, or at least reduced to the problem of doing an integral. One of these (which we have met a couple of times in this chapter) is the motion of a one-dimensional particle subject to a force that depends only on the velocity v, that is, F = F(v). Write down Newton's second law and separate the variables by rewriting it as m dv/F(v) = dt. Now integrate both sides of this equation and show that t = m v dv' F(v¹) Vo Provided you can do the integral, this gives t as a function of v. You can then solve to give v as a function of t. Use this method to solve the special case that F(v) = Fo, a constant, and comment on your result. This method of separation of variables is used again in Problems 2.8 and 2.9.