Question regarding Simple Harmonic Motion...

sankalpmittal

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...

gneill

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

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

sankalpmittal

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

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

sankalpmittal

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

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

A better link to consider might be: http://en.wikipedia.org/wiki/Lisajous .

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