Easy Simple Harmonic Motion Problem

  • #1
150
2

Homework Statement


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?

Homework Equations


x=a*cos(wt+ø)
T=2π/w

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
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,909

Homework Statement


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?

Homework Equations


x=a*cos(wt+ø)
T=2π/w

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
"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?
 
  • Like
Likes i_hate_math
  • #3
150
2
"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?
Then x(t)=1.5 this means cos(¢)=1
 
  • #4
150
2
Then x(t)=1.5 this means cos(¢)=1
pls ignore my previous post,
x(0)=0=1.5cos(0+pi/2)
 
  • #5
ehild
Homework Helper
15,543
1,909
pls ignore my previous post,
x(0)=0=1.5cos(0+pi/2)
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?
 
  • Like
Likes i_hate_math
  • #6
150
2
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?
I see... 0.75 it is
 
  • #7
ehild
Homework Helper
15,543
1,909
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?
 
  • Like
Likes i_hate_math
  • #8
150
2
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?
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
ehild
Homework Helper
15,543
1,909
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)
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.
 
  • Like
Likes i_hate_math
  • #10
150
2
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.
okay so at one end x=-0.75=0.75sin(wt), wt=3π/2
 
  • #11
ehild
Homework Helper
15,543
1,909
Well, where are both particles 0.46 s later? How far are they?
 
  • Like
Likes i_hate_math
  • #12
150
2
Well, where are both particles 0.46 s later? How far are they?
X=sin(0.46w)=0.937282
 
  • #13
ehild
Homework Helper
15,543
1,909
X=sin(0.46w)=0.937282
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
150
2
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
ehild
Homework Helper
15,543
1,909
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)
Yes, and what is the position of the other particle at the same time? What is the distance between them?
 
  • Like
Likes i_hate_math
  • #16
150
2
Yes, and what is the position of the other particle at the same time? What is the distance between them?
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
ehild
Homework Helper
15,543
1,909
Yes, but you forgot the amplitude from both x-s.
Calculate the distance.
 
  • Like
Likes i_hate_math
  • #18
150
2
Yes, but you forgot the amplitude from both x-s.
Calculate the distance.
oh yes it should be
d=0.75sin(3π/2+0.46w+π/5)-0.75sin(3π/2+0.46w)=0.363261895
 
  • #19
ehild
Homework Helper
15,543
1,909
oh yes it should be
d=0.75sin(3π/2+0.46w+π/5)-0.75sin(3π/2+0.46w)=0.363261895
OK, it is correct.
Now question b) :)
 

Related Threads on Easy Simple Harmonic Motion Problem

Replies
5
Views
35K
Replies
1
Views
1K
Replies
7
Views
7K
Replies
2
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
5
Views
568
Replies
5
Views
3K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
995
Top