# Easy Simple Harmonic Motion Problem

1. Apr 29, 2016

### i_hate_math

1. The problem statement, all variables and given/known data
Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.5 m. Each particle has a period of 1.5 s, but they differ in phase by π/5 rad. (a) How far apart are they 0.46 s after the lagging particle leaves one end of the path? (b) Are they then moving in the same direction, toward each other, or away from each other?

2. Relevant equations
x=a*cos(wt+ø)
T=2π/w
3. The attempt at a solution
the amplitude is given by a=1.5m
and T=1.5s
then we have w=2π/T=4.18879..
the two particles are identical except they differ in phase by π/5
my attempt in finding the distance between them after t=0.46s is
d=1.5cos(wt)-1.5cos(wt+π/5)=0.28647
which does not match the correct solution
can someone point out where I went wrong

2. Apr 30, 2016

### ehild

"How far apart are they 0.46 s after the lagging particle leaves one end of the path?"
At t=0, the lagging particle is just leaving one end of the path, say the left end, where x1=0, and turning to move toward the other end at x2=1.5 m. What is the function x(t) then?

3. Apr 30, 2016

### i_hate_math

Then x(t)=1.5 this means cos(¢)=1

4. Apr 30, 2016

### i_hate_math

pls ignore my previous post,
x(0)=0=1.5cos(0+pi/2)

5. Apr 30, 2016

### ehild

It is "simple harmonic motion along a common straight-line segment of length 1.5 m." That can mean from 0 to 1.5, for example, or from -0.75 to 0.75, as an other example. What is the amplitude?

6. Apr 30, 2016

### i_hate_math

I see... 0.75 it is

7. Apr 30, 2016

### ehild

Yes. And what is the x(t) function, so as it is at the left end of the interval at t=0 and the velocity just turning to positive?

8. Apr 30, 2016

### i_hate_math

So for when its at the left end where x=0 at t=0,
x(0)=0.75cos(π/2) ===> ø=π/2
x(t)=0.75cos(wt+π/2)=0.75sin(wt)
and the other particle has
x'(t)=0.75cos(wt+π/5+π/2)=sin(wt+π/5)

9. Apr 30, 2016

### ehild

x(t) = 0.75sin(wt) varies between -0.75 and 0.75, and at t=0 it is at x=0. It is not one end of the line segment.
Decide where are the end of the light segment first.

10. Apr 30, 2016

### i_hate_math

okay so at one end x=-0.75=0.75sin(wt), wt=3π/2

11. Apr 30, 2016

### ehild

Well, where are both particles 0.46 s later? How far are they?

12. Apr 30, 2016

### i_hate_math

X=sin(0.46w)=0.937282

13. Apr 30, 2016

### ehild

The question was "How far apart are they 0.46 s after the lagging particle leaves one end of the path?". The lagging particle leaves the left end when wt = 3π/2. Where is it 0.46 s later?

14. May 1, 2016

### i_hate_math

when it leaves the left end wt=3π/2 ==> t=3π/2w
0.46s later t'=3π/2w + 0.46
the position is x=sin(wt')=sin(3π/2+0.46w)

15. May 1, 2016

### ehild

Yes, and what is the position of the other particle at the same time? What is the distance between them?

16. May 1, 2016

### i_hate_math

the other particle has position x=sin(3π/2+0.46w+π/5) since they differ in phase by π/5
the distance would be the difference between the two x values

17. May 1, 2016

### ehild

Yes, but you forgot the amplitude from both x-s.
Calculate the distance.

18. May 1, 2016

### i_hate_math

oh yes it should be
d=0.75sin(3π/2+0.46w+π/5)-0.75sin(3π/2+0.46w)=0.363261895

19. May 1, 2016

### ehild

OK, it is correct.
Now question b) :)