How long does it take for the box to move from -2.2 m to +2.2 m?

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Homework Help Overview

The problem involves determining the time it takes for a box to move from a position of -2.2 m to +2.2 m, given its position as a function of time described by the equation x = 4.4 m * cos(29/sec * t).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of arccosine to find the corresponding angles for the box's positions and the implications of phase changes over time. There are attempts to reconcile the results with the expected values and significant figures.

Discussion Status

Some participants have provided insights into the phase relationships and the need to consider multiple angles. There is an ongoing exploration of the calculations and interpretations of the results, with no clear consensus yet on the correct approach or answer.

Contextual Notes

Participants note that the problem does not specify which initial phase to use, leading to different interpretations. There is also mention of the potential influence of online grading software on the perceived correctness of answers.

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Homework Statement


The position of the center of the box shown is given by the equation

x = 4.4 m * cos(29/sec * t)

-How long does it take the box to move from -2.2 m to +2.2 m?

Homework Equations


x = 4.4 m * cos(29/sec * t)


The Attempt at a Solution


±.5=cos29t
[arccos(-.5) - arcccos(.5)]/29=.03611 sec

it says its wrong.
 
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It is astounding how many problems become simpler after you’ve
sketched a graph. Also, until you’ve sketched some graphs of functions you really don’t know how they behave.

From Mathematical Tools for Physics

http://img339.imageshack.us/img339/1490/cosj.jpg
 
Last edited by a moderator:
SalsaOnMyTaco said:

The Attempt at a Solution


±.5=cos29t
[arccos(-.5) - arcccos(.5)]/29=.03611 sec

it says its wrong.
I agree with your answer. Perhaps "it" (online grading software?) is expecting fewer significant figures.

azizlwl said:
It is astounding how many problems become simpler after you’ve
sketched a graph. Also, until you’ve sketched some graphs of functions you really don’t know how they behave.
Okay, but your graph also shows agreement with the OP's answer:
From Mathematical Tools for Physics

http://img339.imageshack.us/img339/1490/cosj.jpg
 
Last edited by a moderator:
SalsaOnMyTaco said:
x = 4.4 m * cos(29/sec * t)

-How long does it take the box to move from -2.2 m to +2.2 m?

Homework Equations


x = 4.4 m * cos(29/sec * t)


The Attempt at a Solution


±.5=cos29t
[arccos(-.5) - arcccos(.5)]/29=.03611 sec

it says its wrong.

The box moves from x=-2.2 to x=2.2 . When x=-2.2 the phase is 2.094 or 4.189. At x=2.2, arccos (0.5) = 1.047, but the phase should increase with time so you need to take the angle next to 2.094 or 4.189 with cosine equal to 0.5: It is 2pi-1.047=5.236. (Draw the unit circle to visualize it). You get two possible values for the time: try the other one.

ehild
 
ehild said:
The box moves from x=-2.2 to x=2.2 . When x=-2.2 the phase is 2.094 or 4.189. At x=2.2, arccos (0.5) = 1.047, but the phase should increase with time so you need to take the angle next to 2.094 or 4.189 with cosine equal to 0.5: It is 2pi-1.047=5.236. (Draw the unit circle to visualize it). You get two possible values for the time: try the other one.

ehild

I don't quite understand this step, could you please type a more detailed explanation?
 
arccos gives angles between 0 and pi. cos(wt)= 0.5 corresponds to ωt=1.047 + k*2pi or ωt=-1.047 + m*2pi. The phase ωt increases with time, the final phase has to be greater than the initial one. And you need to subtract the initial phase from the final one to get the time.

The initial phase can be either 2.094 or 2pi-2.094=4.189. The problem does not specify which one. The final phase must be greater then 2.094: 2pi-1.047=5.236. I the first case, the time is Δt=(5.236-2.094)/29=0.108s. In the second case Δt=(5.236-4.189)/29=0.036 s, the same you got.

ehild
 

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