Forced Oscillations (proving diff equation by subsitution)

AI Thread Summary
The discussion revolves around solving a differential equation related to forced oscillations, specifically proving its validity for any time t. The user successfully demonstrated the equation's correctness for t = -phi/w but struggles with the case when t = (pi/2 - phi)/w, particularly in eliminating the variable k. They note that substituting sin = 0 simplifies the problem, but substituting cos = 0 complicates it. Clarifications are sought regarding the natural frequency and the relationship between variables. The conversation emphasizes the need to derive equations for A and phi based on the parameters of the driving force, w, and F_nat.
hjr
Messages
8
Reaction score
0

Homework Statement


By substituting the proper equations I showed that the equation is right when time = phi/w.
Now when I make cos = o and sin = 1 and time = (pi/2 - phi)/w I can't solve the equation.


Homework Equations


If you need to see all the equations i can give it to you but I am pretty sure at this stage they are not needed. The original diff equation is:

m*dx^2 + b* dx + kx = F_nat*cos(wt)


with x equal to:

Asin(wt + phi)

A = F-nat/ mh where h = \sqrt{(\omega^{2}-\omega_{0}^{2})^{2} + b^{2}\omega^{2}/m^{2}}

tan(\phi) = \frac{ (\omega^{2}-\omega_{0}^{2})}{\omega(b/m)}

The Attempt at a Solution



when time = (pi/2 - phi)/w

i got:

F-nat/m * (k-mw^2) = f-nat * (w-nat^2 - w^2)

in my book in a different example you can solve for k and get mw^2 but then that side will be zero. But then the natural frequency has to equal the external forces frequency to make that side zero. I just need a hint. Sorry if this is all a mess. If you need anything clarify I will try my best to do it.
 
Last edited:
Physics news on Phys.org
I am not too entirely sure as to what you are looking to find. Could you please clarify a bit?
 
The solution is x=Asin(wt + phi) if A and phi have the appropriate values which depend on the parameters of the diving force, w and F_nat. You can determine these values by substituting x(t) and its derivatives into the differential equation.This equation has to be true for all values of t. Substitute t=-phi/w, (it is minus!) you get a relation between A, phi and w and F_nat. Substitute t=pi/2-phi/w, you get a second equation. You can solve this system of equations for A and phi in terms of w and F_nat.

ehild
 
Yeah i should of been more clearer, A and phi were already given. I have to prove that the equation is valid for any time t. I did it for t = -phi/w which I was able to show/prove that it does work. Now when i do t=(pi/2-phi/w) i get the solution above but I can't figure out how to get rid of k. The only solution I could think of was that it must be zero on both sides. I am going to edit the question to make it clearer.
 
yea so these are the equations i was given. When i make sin = 0 i can make it work but when cos = o i can't figure out how to get rid of k.
 
What do you mean on w-nat? Or is it w0?

As for k, you know that w02=k/m. If you mean f-nat the same as F-nat, the two sides of the equation are equivalent .ehild
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top