Vector of a particle moving in space help

  • Thread starter zorro
  • Start date
  • #1
1,384
0

Homework Statement



The position vector of a particle moving in space is given by
[tex]\vec{r}[/tex]=[tex](1+cos2wt)[/tex][tex]\hat{i}[/tex] [tex]+[/tex] [tex]3sin^{2}wt[/tex][tex]\hat{j}[/tex][tex]+[/tex][tex]3t[/tex][tex]\hat{k}[/tex]
in the ground frame. All the units are in SI. Find the position vector and amplitude of the particle in a frame S moving along the positive z-axis with a velocity 3m/s. Is the line of S.H.M. of the particle parallel to [tex]4/5[/tex][tex]\hat{i}-3/5[/tex][tex]\hat{j}[/tex]?

Attempt:

The P.V. of the frame S at any time t is 3t [tex]\hat{k}[/tex].
From this we can find out the P.V. of the particle w.r.t. the frame S.

I differentiated it w.r.t. time to find out the velocity vector and found Vmax, which is 5w.
Vmax=Aw
so A=5m

Unfortunately, I got both of them incorrect :/
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,913


Is it the formula?

[tex]\vec {r}=(1+\cos(2\omega t)\hat {i }+3\sin^2(\omega t)\hat {j}+3t\hat{k}[/tex],

(click on it and see the Latex code).

How did you get the result vmax=5w?

The amplitude can be found without the velocity. Find the direction of the SHM and write it along this direction as A(cos2wt)*(unit vector).


ehild
 
Last edited:
  • #3
1,384
0


Correction- The component of the P.V. along x-axis is 1+2cos2wt.
First I found out the velocity, then differentiated it to find the condition of Vmax and got wt = pi/4 (one of the solutions)
Subsitituting the value of wt in the equation for velocity, I found out the magnitude of Vmax as 5w.

How to find out the direction of SHM?
 
  • #4
ehild
Homework Helper
15,543
1,913


The velocity is a vector. Only the magnitude of a vector has maximum or minimum.
A simple harmonic motion always happens along a certain direction. On a table, it is horizontal, a body suspended on a spring oscillates vertically, a body on a slope, connected to a spring oscillates along the slope. The amplitude includes information about the direction of oscillation. In general, it is a vector. Think of the polarization of electromagnetic waves.

Collect the cosine terms: you get a cosine multiplied by a vector: it is the vector amplitude.

ehild
 
  • #5
1,384
0


The velocity is a vector. Only the magnitude of a vector has maximum or minimum.

Magnitude of the vector has max/min when each of its component has max/min. That happens when wt=pi/4.

Collect the cosine terms: you get a cosine multiplied by a vector: it is the vector amplitude.

ehild
I understand that there is a line of motion of any S.H.M. How do I 'collect' the cosine terms? I don't get your point here.
 
  • #6
ehild
Homework Helper
15,543
1,913


Magnitude of the vector has max/min when each of its component has max/min. That happens when wt=pi/4.
Think it over.
I understand that there is a line of motion of any S.H.M. How do I 'collect' the cosine terms? I don't get your point here.
Edit: As sin2(ωt)=0.5(1-cos(2ωt)) both components of the position vector have a time-dependent cos(2ωt) term. Factoring this out, you have the sum of a constant vector and an oscillating one.

ehild
 
Last edited:
  • #7
1,384
0


Think it over.

I was wrong :redface:


As sin2(ωt)=1-cos(2ωt) both components of the position vector have a time-dependent cos(2ωt) term. Factoring this out, you have the sum of a constant vector and an oscillating one.

ehild

I got the oscillating term as (i-3j/2)cos2wt
So the amplitude is (13)1/2/2.
But this is not correct.
 
  • #8
ehild
Homework Helper
15,543
1,913


Well, i made a mistake in the previous post: sin2(ωt)=0.5(1-cos(2ωt)).
You have corrected the first formula in your second post : "Correction- The component of the P.V. along x-axis is 1+2cos2wt."

So the position vector is

(i+3/2j)+(2i+3/2 j) cos(2ωt)

at the end, and the magnitude |(2i+3/2 j)|=5/2.

ehild
 
  • #9
1,384
0


ok, so the line of SHM of particle is 2i + 3/2j which is not parallel to the vector given in the question.

Thanks!
 
  • #10
ehild
Homework Helper
15,543
1,913


ok, so the line of SHM of particle is 2i + 3/2j which is not parallel to the vector given in the question.

Thanks!

Oh my! one more mistake. I am tired of gardening :) It is

2i -3/2j. It is parallel with the given vector.

ehild
 

Related Threads on Vector of a particle moving in space help

Replies
4
Views
5K
Replies
1
Views
6K
  • Last Post
2
Replies
34
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
9
Views
3K
Replies
4
Views
1K
  • Last Post
Replies
8
Views
905
  • Last Post
Replies
4
Views
1K
Top