# Question regarding Simple Harmonic Motion...

by sankalpmittal
Tags: harmonic, motion, simple
 P: 694 1. The problem statement, all variables and given/known data The two linear simple harmonic motions of equal amplitudes , and angular frequencies ω and 2ω are imposed on a particle along the axes of X and Y respectively. If the initial phase difference between them is π/2 , then find the resultant path followed by the particle. 2. Relevant equations http://en.wikipedia.org/wiki/Simple_harmonic_motion 3. The attempt at a solution I tried solving question using x= a sinωt And other is y=a sin(2ωt + π/2) Solving them , I get the wrong answer in terms of x and y.... Please help !! Thanks in advance...
 Mentor P: 10,760 Can you show how you went about "solving them"? We need to see your attempt if we're to know how to help.
 HW Helper Thanks P: 9,265 Hi sankalpmittal, Show please what you have tried? x=sin(ωt) and y=sin(2ωt+π/2) =cos(2ωt) is the parametric representation of a Lissajous curve. See http://en.wikipedia.org/wiki/Lissajous_curve Try to eliminate the ωt term. ehild
P: 694

## Question regarding Simple Harmonic Motion...

 Quote by ehild Hi sankalpmittal, Show please what you have tried? x=sin(ωt) and y=sin(2ωt+π/2) =cos(2ωt) is the parametric representation of a Lissajous curve. See http://en.wikipedia.org/wiki/Lissajous_curve Try to eliminate the ωt term. ehild
Firstly I solved this question using :
a : Amplitude
x=a cosωt
y=a sin(2ωt+π/2)
y=a cos (2ωt)
y= a( cos2ωt-sin2ωt)
y= a(2cos2ωt-1)
y= a{(2x2/a2)-1)

I noticed that this does not match the answer given in my textbook...

Then I made a second attempt :
x=a sinωt
y=a sin (2ωt+π/2)
y=a cos(2ωt)
y= a(1-2sin2ωt)
y= a{1-(2x2/a2)}

Again this does also not match with the answer in my textbook....

Then I made the third attempt :

x= a cosωt
y= a cos(2ωt+π/2)
y= -asin(2ωt)
y= -2a sin(ωt)cos(ωt)
y=-2a √(1-sin2ωt) (x/a)
y=-2a √{1-(x2/a2)} (x/a)
On squaring both sides and simplifying , I got :

y= 4x2{1-(x2/a2)}

I just wanted to know , where I went wrong in my first and second attempt....
HW Helper
Thanks
P: 9,265
 Quote by sankalpmittal Firstly I solved this question using : a : Amplitude x=a cosωt y=a sin(2ωt+π/2) y=a cos (2ωt)
They are ins phase, instead of being shifted by pi/2.

 Quote by sankalpmittal Then I made a second attempt : x=a sinωt y=a sin (2ωt+π/2) y=a cos(2ωt) y= a(1-2sin2ωt) y= a{1-(2x2/a2)} Again this does also not match with the answer in my textbook...
.

That is correct, if x and y are as you assumed.

 Quote by sankalpmittal Then I made the third attempt : x= a cosωt y= a cos(2ωt+π/2) y= -asin(2ωt) y= -2a sin(ωt)cos(ωt) y=-2a √(1-sin2ωt) (x/a) y=-2a √{1-(x2/a2)} (x/a) On squaring both sides and simplifying , I got : y= 4x2{1-(x2/a2)} This answer matched with the answer given in my textbook....
But this answer is wrong as y is not squared. It was correct before squaring.

If you add two perpendicular vibrations you can have different y(x) curves, depending on the initial phases.

ehild
P: 694
 Quote by ehild They are ins phase, instead of being shifted by pi/2. . That is correct, if x and y are as you assumed. But this answer is wrong as y is not squared. It was correct before squaring. If you add two perpendicular vibrations you can have different y(x) curves, depending on the initial phases. ehild
OK , so there was a typo. The answer I got in my third attempt was y2=4x2{1-(x2/a2)} , which matched with the answer given in my textbook. But unfortunately the answer which I got in my first and second attempt , did not match with that given in my textbook.

According to you , both the equations , in my second and third attempt are correct , yet they are different and only latter match with the answer in my textbook. How ?

Also what do you mean by "in phase and not shifted by pi/2" in my first attempt ? Can you explain a bit comprehensively ?
 HW Helper Thanks P: 9,265 Your first attempt was x=a cosωt y=a sin(2ωt+π/2) which is equivalent to y=a cos (2ωt)I do not see any initial phase difference between x and y. Your second attempt is true and so is the third one. Phase difference between x and y is not a clear concept. You get x,y curves of different form if x=sin(ωt), y=cos(2ωt) (y=1-2x2) and when x=cos(ωt) and y=sin(2ωt) (y2=4x2(1-x2)) ehild Attached Thumbnails
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