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10 Nov 2019

Hello. There is one correction to the table. The period(T) needs to be divided by 10. So for m1 the period (T) wouldbe 3.3, m2 would be 4.7. Sorry for the confusion. Theinstructions for this table are below:

Using your measured values, calculate the frequency (Ï) ofeach oscillation, and deduce the value for the spring constant,k.

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Patrina SchowalterLv2

14 Apr 2019

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You see two carts connected by a spring and oscillating on ahorizontal air track. You happen to know that the spring constantof the spring is k = 50.0 N/m. You use Newton's second law todetermine that the period of oscillation is the same for each cartand is related to the masses of the carts by

T = 2Ïâ[ (m₁m₂)/(k(m₁ +m₂))].

Using a position vector, you determine that the positions of thecarts as functions of time can be expressed as
x₁(t) = 2.70[1 - cos(18.0t)] and
x₂(t) = 10.70 + 1.29cos(18.0t)
wher the coordinates are in centimeters and the time is inseconds.

a) Use the principle of conservation of momentum to find the massesof the carts.

b) Generate a table of the values of x₁,x₂ for every 1 s from t = 0 to t = 35 s.
For each of these times, also calculate the position of the centerof mass, the total momentum of the two-cart system, and the forceof the spring on each cart.
Verify that the center of mass does not move, that the totalmomentum is conserved, and that the forces on the carts from thespring have the same magnitude but in opposite directions.

c) Use the table to find the equilibrium length of thespring.

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.

NOTE, you may need to save images to desktop to view in a largerscale.

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