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

  • Thread starter Thread starter SalsaOnMyTaco
  • Start date Start date
  • Tags Tags
    Oscillator
Click For Summary
SUMMARY

The discussion centers on 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). The correct approach involves determining the phase angles corresponding to these positions, specifically 2.094 radians for -2.2 m and 5.236 radians for +2.2 m. The time intervals are calculated as Δt = (5.236 - 2.094)/29 = 0.108 seconds and Δt = (5.236 - 4.189)/29 = 0.036 seconds, confirming the initial calculation was correct but possibly rejected due to significant figures.

PREREQUISITES
  • Understanding of trigonometric functions and their properties
  • Familiarity with phase angles in oscillatory motion
  • Knowledge of inverse trigonometric functions, specifically arccos
  • Ability to interpret and sketch graphs of trigonometric functions
NEXT STEPS
  • Study the unit circle and its application in determining phase angles
  • Learn about significant figures and their impact on calculations in physics
  • Explore the concept of harmonic motion and its mathematical representation
  • Practice solving similar problems involving cosine functions in oscillatory systems
USEFUL FOR

Students studying physics, particularly those focusing on oscillatory motion, as well as educators looking for examples of trigonometric applications in real-world scenarios.

SalsaOnMyTaco
Messages
32
Reaction score
0

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.
 
Physics news on Phys.org
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
 

Similar threads

Replies
16
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
8K
Replies
9
Views
2K
  • · Replies 10 ·
Replies
10
Views
5K
Replies
7
Views
2K
Replies
6
Views
988
  • · Replies 21 ·
Replies
21
Views
2K