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Question regarding Simple Harmonic Motion...

by sankalpmittal
Tags: harmonic, motion, simple
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sankalpmittal
#1
Dec12-12, 10:40 AM
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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...
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gneill
#2
Dec12-12, 11:16 AM
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Can you show how you went about "solving them"? We need to see your attempt if we're to know how to help.
ehild
#3
Dec12-12, 11:17 AM
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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

sankalpmittal
#4
Dec13-12, 03:43 AM
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Question regarding Simple Harmonic Motion...

Quote Quote by ehild View Post
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)}

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....
ehild
#5
Dec13-12, 05:26 AM
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Quote Quote by sankalpmittal View Post
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 Quote by sankalpmittal View Post

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 Quote by sankalpmittal View Post
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
sankalpmittal
#6
Dec15-12, 04:52 AM
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Quote Quote by ehild View Post
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 ?
ehild
#7
Dec15-12, 05:44 AM
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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
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SammyS
#8
Dec15-12, 11:50 AM
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Quote Quote by sankalpmittal View Post
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
...
A better link to consider might be: http://en.wikipedia.org/wiki/Lisajous .

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


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