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

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The discussion revolves around calculating the time it takes for a box to move from -2.2 m to +2.2 m using the equation x = 4.4 m * cos(29/sec * t). Initial attempts yielded a time of approximately 0.036 seconds, which was deemed incorrect, prompting further analysis. Participants emphasized the importance of sketching graphs to understand the behavior of the function better. The correct approach involves determining the initial and final phases of the cosine function, leading to two possible time values: 0.108 seconds and 0.036 seconds. Clarification on phase angles and their relation to time was also discussed, highlighting the need for careful consideration of the problem's parameters.
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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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