PHYS 2004 Lecture 35: Phy 2004

43 views1 pages

ultramarinemammoth89

1 Dec 2017

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

A second order differential equation has two independent solutions 1(x) and 2(x). A theorem states that a general solution of the equation can be written as where a and b are constants whose values are determined by the boundary conditions. If we can find two independent solutions 1(x) and 2(x) by guessing, then this sum is the general solution to the schr dinger equation. Acceptable solutions exist only for discrete values of e. The solutions describe the stationary states of the system. Allow to u(cid:374)dersta(cid:374)d the pri(cid:373)ar(cid:455) features of a parti(cid:272)le"s (cid:373)otio(cid:374) without e(cid:373)phasis o(cid:374) the details. A (cid:373)odel"s su(cid:272)(cid:272)ess is its agree(cid:373)e(cid:374)t with e(cid:454)peri(cid:373)e(cid:374)t. Models need to be tied to measurable quantities. Quantum mechanical models are defined in terms of energies. Requires determining a potential-energy function u(x) that des(cid:272)ri(cid:271)es the parti(cid:272)le"s i(cid:374)tera(cid:272)tio(cid:374)s with the environment.

Related Questions

Learning Goal: Understand how to determine the constants in thegeneral equation for simple harmonic motion, in terms of giveninitial conditions.
A common problem in physics is to match the particular initialconditions - generally given as an initial position x_0 andvelocity v_) at t=0 - once you have obtained the general solution.You have dealt with this problem in kinematics, where theformula

1. x(t) = x_0 + v_0t + at^2

has two arbitrary constants (technically constants of integrationthat arise when finding the position given that the acceleration isa constant). The constants in this case are the initial positionand velocity, so "fitting" the general solution to the initialconditions is very simple.

For simple harmonic motion, it is more difficult to fit the initialconditions, which we take to be

x_0, the position of the oscillator at t = 0, and
v_0, the velocity of the oscillator at t = 0.

There are two common forms for the general solution for theposition of a harmonic oscillator as a function of time t:

2. x(t) = A\cos(omega t + phi) and
3. x(t) = C\cos(omega t) + S\sin(omega t),

where A, phi, C, and S are constants, omega is the oscillationfrequency, and t is time.

Although both expressions have two arbitrary constants--parametersthat can be adjusted to fit the solution to the initialconditions--Equation 3 is much easier to use to accommodate x_0 andv_0. (Equation 2 would be appropriate if the initial conditionswere specified as the total energy and the time of the first zerocrossing, for example.)

Find C and S in terms of the initial position and velocity of theoscillator.
Give your answers in terms of x_0, v_0, and omega. Separate youranswers with a comma.

Problems that require solving the three-dimensional Schrödinger equation can often be reduced to related one-dimensional problems. An example of this would be the particle in a cubical box.

Consider a cubical box with rigid walls (i.e., outside of the cube) and edges of length L. The general solution for this problem is

, where, n_x, n_y, and n_z are all positive integers. Note that this solution is just the product of three solutions to the one-dimensional particle in a box. The energy corresponding to the three-dimensional solution is just the sum of the energies for each of the three one-dimensional solutions:

Part A

What is the smallest allowed energy E0 for a particle in a cubical box?

Express your answer in terms of , m, and L.

Part B

If you are dealing with a very large n_rs, you can assume that each state (point with integer coordinates) corresponds roughly to one unit of volume inside of the sphere. So, the number of states is approximately equal to the volume of the octant of the sphere. Use this idea to find the number N of states with energy less than or equal to for a large n_rs.

Express your answer in terms of n_rs.

scarletduck111

which of the vectors below best represents the direction of the impulse vector j⃗ ?

To learn about the impulse-momentum theorem and its applications in some common cases.

Using the concept of momentum, Newton's second law can be rewritten as

?F? =dp? dt, (1)

where ?F? is the net force F? net acting on the object, and dp? dt is the rate at which the object's momentum is changing.

If the object is observed during an interval of time between times t1 and t2, then integration of both sides of equation (1) gives

?t2t1?F? dt=?t2t1dp? dtdt. (2)

The right side of equation (2) is simply the change in the object's momentum p2??p1?. The left side is called the impulse of the net force and is denoted by J? . Then equation (2) can be rewritten as

J? =p2??p1?.

This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force F? net acting along the direction of motion, the impulse-momentum theorem can be written as

F(t2?t1)=mv2?mv1. (3)

Here F, v1, and v2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.

( The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x-axis. The net force F is acting on the particle along the x-axis. F is a constant force.)

A) The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0.

B) The particle starts from rest at t=0. What is the magnitude v of the velocity of the particle at time t? Assume that t>0.

C) The particle has the momentum of magnitude p1 at a certain instant. What is p2, the magnitude of its momentum ?t seconds later?

D) The particle has the momentum of magnitude p1 at a certain instant. What is v2, the magnitude of its velocity ?t seconds later?

E) Find the magnitude of the impulse J delivered to the particle.

F) Which of the vectors below best represents the direction of the impulse vector J??
(Figure 1)

1
2
3
4
5
6
7
8

puceworm982

PHYS 2004 Lecture 35: Phy 2004

Document Summary

Get access

Related textbook solutions

Physics: Principles with Applications

Related Documents

PHYS 2004 Lecture Notes - Lecture 34: Matter Wave

Physics 2102A/B Lecture 8: Ch6-Schrodinger_Equation

CH202 Study Guide - Final Guide: Matter Wave, Partial Differential Equation, Bohr Model

Related Questions