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Question regarding Simple Harmonic Motion... 
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#1
Dec1212, 10:40 AM

P: 758

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... 


#2
Dec1212, 11:16 AM

Mentor
P: 11,678

Can you show how you went about "solving them"? We need to see your attempt if we're to know how to help.



#3
Dec1212, 11:17 AM

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P: 10,547

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 


#4
Dec1312, 03:43 AM

P: 758

Question regarding Simple Harmonic Motion...
a : Amplitude x=a cosωt y=a sin(2ωt+π/2) y=a cos (2ωt) y= a( cos^{2}ωtsin^{2}ωt) y= a(2cos^{2}ωt1) y= a{(2x^{2}/a^{2})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(12sin^{2}ωt) y= a{1(2x^{2}/a^{2})} 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 √(1sin^{2}ωt) (x/a) y=2a √{1(x^{2}/a^{2})} (x/a) On squaring both sides and simplifying , I got : y= 4x^{2}{1(x^{2}/a^{2})} This answer matched with the answer given in my textbook.... I just wanted to know , where I went wrong in my first and second attempt.... 


#5
Dec1312, 05:26 AM

HW Helper
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P: 10,547

That is correct, if x and y are as you assumed. If you add two perpendicular vibrations you can have different y(x) curves, depending on the initial phases. ehild 


#6
Dec1512, 04:52 AM

P: 758

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 ? 


#7
Dec1512, 05:44 AM

HW Helper
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P: 10,547

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=12x^{2}) and when x=cos(ωt) and y=sin(2ωt) (y^{2}=4x^{2}(1x^{2})) ehild 


#8
Dec1512, 11:50 AM

Emeritus
Sci Advisor
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P: 7,808

Also see: http://en.wikipedia.org/wiki/Lemniscate_of_Gerono . 


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