Mass-Spring System Beats and Resonance

In summary: I do not know that equation we put XP on the original equation, I could not solveNow 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 xp into md2x/dt2 + kx = F0cosωt, to find what A and B...I do not understand how we find the particular solution of the second one.We are trying to solve the equation of md2x/dt2 + kx = F0cosωt. Since the general solution is x(t) = Acosωt + Bsinωt, we set the particular solution to be x(t) = t(Acosωt
  • #1
viciado123
54
0
Beats and Resonance


In the Beat not have friction force, correct ?

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

We can write as

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

If [tex] w \not= w_o[/tex]

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

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

And why find that: [tex]x_p[/tex] is [tex]x_p = \frac{F_o}{m(w_o^2 -w^2)}cos(wt)[/tex] ?
 
Physics news on Phys.org
  • #2
viciado123 said:
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.

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

We can write as

[tex]\frac{d^2x}{dt^2} + w_o^2 x = \frac{F_o}{m} cos(wt)[/tex]
Defining [itex]w_0[/itex] to be a square root of k, yes.

If [tex] w \not= w_o[/tex]
I suspect you will soon learn what to do in case those are equal!

Assuming (Particular solution)
[tex]x_p = acos(wt) + bsin(wt)[/tex] 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 [tex]a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt)[/tex] ?
If y= a cos(w t)+ b sin(wt), then y'= -aw sin(wt)+ bw cos(wt) and [itex]y"= -aw^2 cos(wt)- bw^2 sin(wt)[/itex]. Put those formulas for y" and y into the differential equation.
And why find that: [tex]x_p[/tex] is [tex]x_p = \frac{F_o}{m(w_o^2 -w^2)}cos(wt)[/tex] ?
The equation you get from the above,
[tex]a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt)[/tex]
must be true for all t. In particular, at t= 0, we must have
[tex]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]
which says that
[tex]a= \frac{F_0}{m(w_0^2- w^2)}[/tex]

and taking [itex]t= \pi/2[/itex], we must have
[tex]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[/tex]
so that b= 0.
 
  • #3
Why we start this equation [tex]a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt)[/tex] ?
Where does it come from?
 
  • #4
viciado123 said:
Assuming (Particular solution)
[tex]x_p = acos(wt) + bsin(wt)[/tex] Why we have assuming this ?

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

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

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! :wink:), which in this case is the given xp.

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

viciado123 said:
Why we start this equation [tex]a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt)[/tex] ?
Where does it come from?
that equation! :rolleyes:.
 
  • #5
Ok. Thanks.

The solution we have is:

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

From this equation how find the solution:
[tex]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}) [/tex] ?

I think when [tex]w[/tex] is near [tex]w_o[/tex] produces beat, movement with varying amplitude
 
  • #6
(type "\left(" and "\right)" for big brackets, and have an omega: ω :wink:)

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

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

viciado123 said:
I think when [tex]w[/tex] is near [tex]w_o[/tex] produces beat, movement with varying amplitude
Try putting ω = ω0 + λ, where λ is small. :wink:
 
  • #7
tiny-tim said:
(type "\left(" and "\right)" for big brackets, and have an omega: ω :wink:)

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. :wink:

Thanks. Beat it is.

About Resonance, need [tex]w = w_o[/tex]

[tex]x(t) = Acos(w_o t) + Bsin(w_o t)[/tex]

Where the particular solution

[tex]x_p(t) = t(Acos(w_ot)+Bsin(w_ot))[/tex]

How I find the solution: => [tex]x_p(t) = \frac{F_o}{2mw_o}tsin(w_ot)[/tex] ?
 
  • #8
Same as before … you put that xp into the original equation, and differentiate where appropriate. :smile:
 
  • #9
tiny-tim said:
Same as before … you put that xp into the original equation, and differentiate where appropriate. :smile:

I do not know that equation we put XP on the original equation, I could not solve
 
  • #10
viciado123 said:
Where the particular solution

[tex]x_p(t) = t(Acos(w_ot)+Bsin(w_ot))[/tex]

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

viciado123 said:
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 xp into md2x/dt2 + kx = F0cosωt, to find what A and B are.
 
  • #11
tiny-tim said:
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 xp into md2x/dt2 + kx = F0cosωt, to find what A and B are.

Derivative great. I do after the with a program :smile:

The general solution is

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

Is corret ?

A and B determine to initial conditions again ?
 
  • #12
Yes. :smile:
 
  • #13
tiny-tim said:
Yes. :smile:

Thank you
 
  • #14
Someone has a graphic example of the system to beat and resonance?
 
  • #15
This system [tex]w_o[/tex] is the frequency of oscillation of the system and [tex]w[/tex] is frequency of oscillation of the external force ?
 
  • #16
tiny-tim said:
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! :wink:), .

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.
 

Related to Mass-Spring System Beats and Resonance

What is a mass-spring system?

A mass-spring system is a physical system that consists of a mass attached to a spring. The mass is able to oscillate back and forth due to the force of the spring.

What causes beats in a mass-spring system?

Beats in a mass-spring system are caused by the interference of two waves with slightly different frequencies. This results in a periodic variation in the amplitude of the resulting wave.

How does resonance occur in a mass-spring system?

Resonance in a mass-spring system occurs when the natural frequency of the system matches the frequency of the external force being applied. This causes a dramatic increase in the amplitude of the oscillations.

What are the applications of mass-spring systems?

Mass-spring systems have a variety of applications, including in musical instruments, shock absorbers, and seismometers. They are also used in engineering and physics experiments to study oscillatory motion.

How can the resonance frequency of a mass-spring system be altered?

The resonance frequency of a mass-spring system can be altered by changing the mass or stiffness of the spring. The frequency can also be altered by changing the external force being applied to the system.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
989
  • Introductory Physics Homework Help
Replies
14
Views
1K
Replies
1
Views
622
  • Introductory Physics Homework Help
Replies
17
Views
587
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Differential Equations
Replies
7
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
2K
  • MATLAB, Maple, Mathematica, LaTeX
Replies
1
Views
3K
  • Differential Equations
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
430
Back
Top