we will firstly notice the undamped case the

Question

We will firstly notice the undamped case. The differential equation under this case is,

mu'' + ku  = F(t)

It is just a non-homogeneous differential equation and we identify how to resolve these. The general solution will be,

u(t) = u_c(t) + U_p(t)

Here the complementary solution is the solution to the free, undamped vibration case. To determine the particular solution we can use either undetermined coefficients or variation of parameters depending on that we determine easier for a specified forcing function.

There is an exact type of forcing function which we should take a look at as this leads to some interesting results. Let's assume that the forcing function is a simple periodic function of the type,

F(t) = F₀ cos(wt)           or F(t) = F₀ sin(wt)

For the reasons of this discussion we'll utilize the first one. By using this, the IVP turns into,

mu'' + ku = F₀ cos(wt)

As pointed out above, the complementary solution, is

u(t) = c₁ cos(w₀t)  + c₂ sin(w₀t)

Here ω₀ is the natural frequency.

We will require being careful in finding an exact solution. The purpose for this will be clear if we utilize undetermined coefficients. Along with undetermined coefficients our guess for the types of the particular solution would be as,

U_P(t) = A cos(w₀t)  + B sin(wt)

Here, this guess will be problems if ω₀= ω. If it were to occur the guess for the particular solution is accurately the complementary solution and so we'd require to add in a t. Obviously if we don't have ω₀= ω then there will be nothing wrong with the guess.

Therefore, we will require looking at this in two cases.

1.      ω₀ ≠ ω

Under this case our initial guess is okay as it won't be the complementary solution. However upon differentiating the guess and plugging this in the differential equation and simplifying we find,

(-m w²A + kA) cos(wt) + (-m w²B + kB) sin(wt) = F₀cos(wt)

Setting coefficients equal provides,

cos(wt):           (-m w² + k)A = F₀         => A= F₀/(k-mw²)

sin(wt):            (-m w² + k)B = 0          => B = 0

So, the particular solution is now,

U_P(t) = (F₀/(k-mw²)) . cos(wt)

= (F₀/m((k/m-)w²)) . cos(wt)

= ((F₀/m(ω²₀ - ω²)) cos(ωt)

Remember that we rearranged things a little. Depending upon the form which you'd like the displacement to be here we can have either of the subsequent.

u(t) = c₁ cos(w₀t)  + c₂ sin(w₀t) +((F₀/m(ω²₀ - ω²)) cos(ωt)

u(t) = R cos(w₀t - d) + ((F₀/m(ω²₀ - ω²)) cos(ωt)

If we utilized the sine form of the forcing function we could determine a same formula.

2. w₀ = w

Under this case we will require to add in a t to the guess for the particular solution.

U_P(t) = At cos(w₀t)  + Bt sin(w₀t)

Remember that we went in front and acknowledge that ω₀ =ω in our guess. Acknowledging this will assist with several simplification which we'll require to do later on. Differentiating our guess, plugging this in the differential equation and simplifying provides us the following:

(-m w₀² + k) At cos(wt) + (-m w₀² + k) Bt sin(wt) + 2mw₀ B cos(wt)- 2mw₀A sin(wt) = F₀ cos(ωt)

Before setting coefficients equivalent, let's notice the definition of the natural frequency and see that,

-m w₀² + k = -m (√(k/m))² + k = -m (k/m) + k = 0

Therefore, the first two terms in fact drop out that is a very good thing and this provides us,

2m w₀ cos(wt) - 2mw₀A sin(wt) = F₀ cos(ωt)

Here setting coefficients equal provides,

cos(wt):           2m w₀ B = F₀    => B= F₀/2mw₀

sin(wt):            2m w₀ A = 0     => A = 0

Under this case exactly will be,

U_P(t) = (F₀/2mw₀) t sin(w₀t)

The displacement for this case is now,

u(t) = c₁ cos(w₀t)  + c₂ sin(w₀t) +((F₀/2mω₀) t sin(ωt))

u(t) = R cos(w₀t - d) + ((F₀/2mω₀) t sin(ωt))

It's totally depending on the form which you prefer for the displacement.

Therefore, what was the point of the two cases now? Fine in the first case, ω₀ ≠ω our displacement function comprises two cosines and is well and nice behaved for all time. In contrast, the second case, ω₀ =ω will contain some serious matters at t raises. The addition of the t in the particular solution will implies that we are going to notice an oscillation that grows in amplitude as t increases. This case is termed as resonance and we would usually like to ignore this at all costs.

In this case resonance arose through assuming that the forcing function was,

F(t) = F₀ cos(wt)

We would also have the opportunity of resonance if we supposed a forcing function of the form.

F(t) = F₀ sin (wt)

We must also take care to not suppose that a forcing function will be in one of these two forms. Forcing functions can approach in a broad variety of forms. If we do run in a forcing function different from the one which used now you will have to go by undetermined coefficients or variation of parameters to find out the particular solution.