Mass-Spring System Beats and Resonance

[viciado123]

Beats and Resonance


In the Beat not have friction force, correct ?

m \frac{d^2x}{dt^2} + kx = F_o cos(wt)

We can write as

\frac{d^2x}{dt^2} + w_o^2 x = \frac{F_o}{m} cos(wt)

If w \not= w_o

Assuming (Particular solution)
x_p = acos(wt) + bsin(wt) Why we have assuming this ?

How find a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt) ?

And why find that: x_p is x_p = \frac{F_o}{m(w_o^2 -w^2)}cos(wt) ?

 

[HallsofIvy]

Beats and Resonance


In the Beat not have friction force, correct ?

No, you can have beats with a friction force- but they die away quickly. In "real life" there is always friction, but you can hear "beats" if you play two guitar strings that are almost tuned to the same note.

m \frac{d^2x}{dt^2} + kx = F_o cos(wt)

We can write as

\frac{d^2x}{dt^2} + w_o^2 x = \frac{F_o}{m} cos(wt)

Defining w_0 to be a square root of k, yes.

If w \not= w_o

I suspect you will soon learn what to do in case those are equal!

Assuming (Particular solution)
x_p = acos(wt) + bsin(wt) Why we have assuming this ?

Because it works! You should have learned, long before this, that the solutions to "linear differential equations with constant coefficients" are only of a few limited kinds of functions:
exponentials, sine and cosine, polynomials, and products of those.

And the method of "undetermined coefficients" for finding "specific solutions" for such equations uses the fact that if the right hand side is one of those kinds of functions, the "specific solution" will also be of that kind, with certain adjustments that you may still be learning about.

Note that this equation has an infinite number of solutions and we are looking for just one. So it doesn't hurt to look specifically for particular types of solutios.

How find a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt) ?

If y= a cos(w t)+ b sin(wt), then y'= -aw sin(wt)+ bw cos(wt) and y"= -aw^2 cos(wt)- bw^2 sin(wt). Put those formulas for y" and y into the differential equation.

And why find that: x_p is x_p = \frac{F_o}{m(w_o^2 -w^2)}cos(wt) ?

The equation you get from the above,
a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt)
must be true for all t. In particular, at t= 0, we must have
a(w_0^2- w^2)(1)+ b(w_0^2- w^2)(0)= a(w_0^2- w^2)= \frac{F_0}{m}(1)[/quote]<br /> which says that <br /> a= \frac{F_0}{m(w_0^2- w^2)}<br /> <br /> and taking t= \pi/2, we must have<br /> a(w_0^2- w^2)(0)+ b(w_0^2- w^2)(1)= b(w_0^2- w^2)= \frac{F_0}{m}(0)= 0<br /> so that b= 0.

 

[viciado123]

Why we start this equation a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt) ?
Where does it come from?

 

[tiny-tim]

Assuming (Particular solution)
x_p = acos(wt) + bsin(wt) Why we have assuming this ?

How find a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt) ?

And why find that: x_p is x_p = \frac{F_o}{m(w_o^2 -w^2)}cos(wt) ?

You don't seem to be grasping how the general-and-particular solution method works.

For the general solution, you pretend the RHS is 0.

For the particular solution, you make an intelligent guess (that's all it is! ), which in this case is the given x_p.

You then put that x_p into the original equation, differentiate where appropriate, and that gives you …

Why we start this equation a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt) ?
Where does it come from?

… that equation! .

 

[viciado123]

Ok. Thanks.

The solution we have is:

x(t) = \frac{F_o}{m(w_o^2 - w^2)} (cos(wt) - cos(w_ot))

From this equation how find the solution:
x(t) = [ \frac{2F_o}{m(w_o^2-w^2)} sin(\frac{(w_o-w)t}{2}) ] sin(\frac{(w_o+w)t}{2}) ?

I think when w is near w_o produces beat, movement with varying amplitude

 

[tiny-tim]

(type "\left(" and "\right)" for big brackets, and have an omega: ω )

That uses one of the trigonometric identities which you need to learn …

cosA - cosB = 2sin((A+B)/2)sin((A-B)/2)​

I think when w is near w_o produces beat, movement with varying amplitude

Try putting ω = ω₀ + λ, where λ is small.

 

[viciado123]

(type "\left(" and "\right)" for big brackets, and have an omega: ω )

That uses one of the trigonometric identities which you need to learn …

cosA - cosB = 2sin((A+B)/2)sin((A-B)/2)​

Try putting ω = ω₀ + λ, where λ is small.

Thanks. Beat it is.

About Resonance, need w = w_o

x(t) = Acos(w_o t) + Bsin(w_o t)

Where the particular solution

x_p(t) = t(Acos(w_ot)+Bsin(w_ot))

How I find the solution: => x_p(t) = \frac{F_o}{2mw_o}tsin(w_ot) ?

 

[tiny-tim]

Same as before … you put that x_p into the original equation, and differentiate where appropriate.

 

[viciado123]

Same as before … you put that x_p into the original equation, and differentiate where appropriate.

I do not know that equation we put XP on the original equation, I could not solve

 

[tiny-tim]

Where the particular solution

x_p(t) = t(Acos(w_ot)+Bsin(w_ot))

How I find the solution: => x_p(t) = \frac{F_o}{2mw_o}tsin(w_ot) ?

I do not know that equation we put XP on the original equation, I could not solve

Now you're worrying me, did you not understand how the particular solution was found in the original case?

In this new case, put the new x_p into md²x/dt² + kx = F₀cosωt, to find what A and B are.

 

[viciado123]

Now you're worrying me, did you not understand how the particular solution was found in the original case?

In this new case, put the new x_p into md²x/dt² + kx = F₀cosωt, to find what A and B are.

Derivative great. I do after the with a program

The general solution is

x(t) = Acos(w_ot) + Bsin(w_ot) + \frac{F_o}{2mw_o}t sin(w_ot)

Is corret ?

A and B determine to initial conditions again ?

 

[tiny-tim]

Yes.

 

[viciado123]

Yes.

Thank you

 

[viciado123]

Someone has a graphic example of the system to beat and resonance?

 

[viciado123]

This system w_o is the frequency of oscillation of the system and w is frequency of oscillation of the external force ?

 

[elibj123]

You don't seem to be grasping how the general-and-particular solution method works.

For the general solution, you pretend the RHS is 0.

For the particular solution, you make an intelligent guess (that's all it is! ), .

Actually there's a general technicall method called the variation of parameters which produces the general solution given any arbitrary "input" function. So it's not just a guess.

In specific cases (where the derivative of the input has its own form) it's much more easy to guess the solution.