LegendCero

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1 years, 257 days

MaplePrimes Activity


These are replies submitted by LegendCero

@tomleslie 

First of all thanks for the reply and for your effort. Oscillator index shows which oscillator has how much energy as time passes and the energy propagates through our system with the form of breathers.  

@Carl Love 

 

3. Yes the total energy of the system is the kinetic energy of x_1 due to its initial velocity.

5. You could try that and check the results.

 

EDIT: Because we are interested (as much as the energy plot is concerned) only on the first oscillations, you could use maybe n=10 instaed of 50?

 

THANKS a lot about your interest and effort.

@Carl Love First of all thanks for the reply. Energy transfers between these two lines of oscillators, one oscillator at a time. For example, at t=0 oscillator x1 has the total energy of the system. Then energy tranfers thgrouh propagating breather to oscillator y2 then x3 etc. That can be seen at the two red-blue graphs. "An impulsive excitation is applied at t = 0, to the leading oscillator of one of the lattices – the ‘‘excited lattice’’ – while the other, ‘‘absorbing lattice’’, is not directly forced. This is equivalent to an initial velocity of the excited oscillator at t = 0 with all other oscillations assumed at rest. We are interested only in primary wave transmission, and not be concerned by reflections at the boundaries."

1. It shows which oscillator "has energy" at a specific time.

2. Yes it is the normalised energy.

3. - This is the only thing I am not sure myself. I try to figure this out but I tihnk is the kinetic energy after some tranformation.

4. The equations are right. For example if we have 6 oscillators then we have 6 equations - mx''1 equation, my''1, mx''2, my''2, mx''3, my''3.

5. Initial positions x1(0) = y1(0) = 0 (for the rest I think we can use some small step h). Initial velocities x'1(0) = v0 = 0.06 m/s , the rest is zero.

6.It is the initial velocity of x1. At t=0 everything is at rest except x1.

7. Yes.

8. n=50 (50 oscilltors per line)

Thanks again. Ask me anything more you want.

@tomleslie Thanks for the reply.

@Rouben Rostamian  Thanks for the reply !

@Rouben Rostamian  Thank you, thank you very much !!!! One question. Line 13. How do you find that the plot has to be x^2/4 ?

@acer I found my mistake. I had another x*y term insde the verify command from another exercise. 

@acer Thanks !!

@Carl Love Thanks!!! No, there isn't any particular reason to use the verify command. I just wanted to try the verify command too.

Thanks for all the answers !!

@Kitonum Thank you very much for the help !

@tomleslie Thank you very much!!

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