# Attributes of an Oscillator

1. Jun 23, 2012

### SalsaOnMyTaco

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
x = 4.4 m * cos(29/sec * t)

3. The attempt at a solution
±.5=cos29t
[arccos(-.5) - arcccos(.5)]/29=.03611 sec

it says its wrong.

2. Jun 23, 2012

### azizlwl

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 [Broken]

Last edited by a moderator: May 6, 2017
3. Jun 24, 2012

### Redbelly98

Staff Emeritus

Last edited by a moderator: May 6, 2017
4. Jun 24, 2012

### ehild

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

5. Jun 24, 2012

### SalsaOnMyTaco

I dont quite understand this step, could you please type a more detailed explanation?

6. Jun 24, 2012

### ehild

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