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23 Nov 2019

A one dimentional harmonic ossilator is changed by the additionof another force so that

V = b/x^2 + 1/2 mÏ_c^2x^2. One resulting wave functionis psi = Ax^ne^{-ax^2} and

d²psi / dx² = [n(n-1)/x^2 - 2a(2n+1) +4a^2x^2] psi

a) evaluate a. b) find E in terms of b, h and Ï_c

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Related questions

A one dimensional harmonic oscillator is changed by the addition of another force so that V= (b/x2) + ½ mω2x2. One resulting wave function is ψ= Ax^ne-ax^2 and d2ψ/dx2 =[n(n-1)/x2– 2a(2n+1) + 4a2x2]ψ. Evaluate a and find E in terms of b, h and ω.

Good Vibes: Introduction to Oscillations

The conditions that lead to simple harmonic motion are as follows:

There must be a position of stable equilibrium.

There must be a restoring force acting on the oscillating object. The direction of this force must always point toward the equilibrium, and its magnitude must be directly proportional to the magnitude of the object's displacement from its equilibrium position. Mathematically, the restoring force F⃗ is given by ⃗, where is the displacement from equilibrium and k is a constant that depends on the properties of the oscillating system.

The resistive forces in the system must be reasonably small.

Consider a block of mass m attached to a spring with force constant k, as shown in the figure(Figure 1) . The spring can be either stretched or compressed. The block slides on a frictionless horizontal surface, as shown. When the spring is relaxed, the block is located at x=0. If the block is pulled to the right a distance A and then released, A will be the amplitude of the resulting oscillations.

Assume that the mechanical energy of the block-spring system remains unchanged in the subsequent motion of the block.

Part A

After the block is released from x=A, it will

(a) remain at rest

(b) move to the left until it reaches equilibrium and stop there.

(d) move to the left until it reaches x=−A and then begin to move to the right.

The time it takes the block to complete one cycle is called the period. Usually, the period is denoted T and is measured in seconds.

The frequency, denoted f, is the number of cycles that are completed per unit of time: f=1/T. In SI units, f is measured in inverse seconds, or hertz (Hz).

Part B

If the period is doubled, the frequency is

(a) unchanged

(b) doubled

Part C

An oscillating object takes 0.10 s to complete one cycle; that is, its period is 0.10 s. What is its frequency f?

Express your answer in hertz.

Part D

If the frequency is 40 Hz, what is the period T?

Express your answer in seconds.

The following questions refer to the figure (Figure 2) that graphically depicts the oscillations of the block on the spring.

Note that the vertical axis represents the x-coordinate of the oscillating object, and the horizontal axis represents time.

Part E

Which points on the x-axis are located at a distance A from the equilibrium position?

(a) R only

(b) Q only

Part F

Suppose that the period is T. Which of the following points on the t-axis are separated by the time interval T?

(a) K and L

(b) K and M

(d) L and N

(e) M and P

Now assume for the remaining Parts G - J, that the x coordinate of point R is 0.12 m and the t coordinate of point K is 0.0050 s.

Part G

What is the period T?

Express your answer in seconds.

Part H

How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement?

Express your answer in seconds.

Part I

What distance d does the object cover during one period of oscillation?

Express your answer in meters.

Part J

What distance d does the object cover between the moments labeled K and N on the graph?

Express your answer in meters.

emeraldyak680

Please refrane from answering if you are unfamiliar with MSExcel and inputing formulas for Cells. Thanks.

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Part b: Collision in Two-Dimensions

The law of conservation of momentum is true for any and everydirection. So, for an isolated system, the momentum in the x, y andz directions or any other direction does not change.

Consider two objects colliding in deep space. Since the two objectstogether can be considered an isolated system, the total momentumof the system (the sum of the momenta of the two objects) isconserved in both the x and y directions. That is

pi-x-total=pf-x-total or pix1+pix2=pfx1+pfx2

pi-y-total=pf-y-total or piy1+piy2=pfy1+pfy2

If we assume object one is moving in the x-direction, object two ismoving in the y-direction and the collision is perfectlyinelastic,
we can rewrite these equations as

m1*vix1=(m1+m2)vfx

m2*viy2=(m1+m2)vfy

These equations can be solved to find the x and y components of theobject's velocity after the collision. The magnitude of vf can befound using the Pythagorean Theorem. The direction of the objectsafter the collision can be found using the inverse cosinefunction

Î¸=cos-1(vfx/|vf|)

Since cos-1 outputs an angle between 0 and Ï radians, we must useÎ¸new=2Ï-Î¸old if vfy is negative to get a correct value. Use thebuilt in function 'if' to find the true value of Î¸. Afterwards,convert the angle into degrees using the built in function'degrees' .

i) Fill in the missing values in the table on the bottom usingformulas to find each value