# Exciting a normal mode of $N$ coupled oscillator with driving force

Suppose we have $$N$$ coupled oscillator with the fixed ends.

We can find the normal modes of this system by considering an infinite system and using space translation symmetry to diagonalize the associated matrix for equation of motion.
In infinite system, we have the solution for the normal for any oscillator as
$$y_j=\mathbb{Re}\Big[e^{i(\omega t+\phi)}(Ae^{ijka}+Be^{-ijka})\Big]\tag{1}$$
with $$\omega^2=2\frac{T}{ma}(1-\cos(ka))$$ where $$k$$ can take any value.

Now when we impose the boundary condition that $$y_0=y_{N+1}=0$$, then we get
$$y_{j,m}=A\sin(jk_ma)\cos(\omega_m t+\phi)\tag{2}$$
where $$k_m=\frac{m\pi}{N+1}$$ and $$m=1,2,..,N$$ are $$N$$ normal modes.
So, now all the frequencies are not allowed in normal mode, it gets discretized.

We know that with external force we can excite the normal mode of a system.
I have a doubt that how to apply the driving force in this system.
Should we have to displace the particle $$0$$ with displacement $$y_o=A_o\sin\omega' t$$, that will drive the other particles?
But if we drive the particle $$0$$ in this way, then from $$(1)$$, we get
$$y_0=(A+B)\cos(\omega t+\phi)=A_o\sin(\omega' t)$$
$$\implies A+B=A_o,\;\phi=3\pi/2\;and\;\omega=\omega'\tag{3}$$
Thus $$\omega'^2=2\omega_o^2(1-\cos(ka))\implies \cos(ka)=\frac{2\omega_o^2-\omega'^2}{2\omega_o^2}$$
Thus, $$\sin(ka)=\frac{\omega'\sqrt{\omega'^2-4\omega_o^2}}{2\omega_o^2}$$
Then we can use $$y_{N+1}=0$$, to find the values for $$A$$ or $$B$$ as a function of driver frequency $$\omega'$$.
$$\implies Ae^{i(N+1)ka}+Be^{-i(N+1)ka}=0$$
$$\implies A_oe^{i(N+1)ka}-B(\sin(N+1)ka)=0\tag{4}$$

Doubt
But from $$(3)$$ is there any way to show that when $$\omega'=\omega_m$$ then the amplitude of the oscillations explodes (resonance)? $$\sin(N+1)ka$$ and $$\cos(N+1)ka$$ has binomial expansion. But I don't know whether this leads to the desired result.
From $$(3)$$, I am not sure whether will we get
$$B(\omega')\;\alpha\;[(\omega'-\omega_1)(\omega'-\omega_2)...(\omega'-\omega_N)]^{-1}$$.
Because only then, the amplitude maximizes when the driver frequency equals the normal mode frequency.

• First you impose the boundary condition $y_0=0$ and then you want to drive it. Without looking further through your question, that is probably what causes your issue. Commented Mar 20, 2023 at 13:39
• @Samuel, thanks for the reply. But can you suggest then how can we excite the normal mode? Because to excite a normal mode we have to give a periodic drive to an oscillator.
– Iti
Commented Mar 20, 2023 at 13:47
• Well if you want to excite exactly one normal mode the driving motion should form an identical standing wave as the corresponding normal mode. Commented Mar 20, 2023 at 14:13
• So, basically if we have 10 resonators with fixed ends and they oscillate such that when they are at maximum distance from the equilibrium position, they form sine wave. So, we have to oscillate the first in the same way and gradually it excites the normal mode. Is that so?
– Iti
Commented Mar 20, 2023 at 15:58
• Also, for two coupled oscillators driven by some periodic force, we can easily solve the differential equation and get that the amplitude of oscillation is maximum when the frequency of the drive matches the normal mode frequency. But is there any way to generalize this result for n- coupled oscillators.
– Iti
Commented Mar 20, 2023 at 16:00

Given: $$\omega^2=2(1-\cos k)$$ (Setting $$a=1$$ and $$T/ma=1$$ up to a change of length and time scale) you want to find the $$\omega$$ poles of: $$A=\frac{1}{\sin(k(N+1))}$$ which is the prefactor of your oscillator amplitudes for $$n=1…n$$: $$y_n=A\sin(k(N+1-n))$$ You anticipate that they will be at the frequencies of the modes where $$n=1…N$$: $$\omega_n^2=2(1-\cos k_n)\\ k_n=\frac{\pi n}{N+1}$$

Actually, it is best to think in terms of $$k\in[0,\pi]$$ rather than in terms of $$\omega\in[0,2]$$. Indeed, both points of view are equivalent since the dispersion relation is bijective in these domains.

You can notice that the $$k_n$$ are zeros of: $$\sin((N+1)k)$$ You therefore obtain the explosion of amplitudes at the eigenmodes.

You can go further by noticing that they are the only zeros in $$[0,\pi]$$ if you do not forget $$k=0,\pi$$ or equivalently $$\omega=0,2$$. But if you consider the full expression of $$y_n$$, you’ll notice that the spatial dependence will cancel these divergences: \begin{align} k&=0 & y_n&= 1-\frac{n}{N+1} \\ k&=\pi & y_n&=(-1)^n\left( 1-\frac{n}{N+1} \right)\\ \end{align} Thus, the spectrum is exactly the places of divergence of the amplitudes as expected.

Small caveat, this assumes that $$\omega\leq 2$$. When $$\omega>2$$, $$k$$ acquires an imaginary part: $$k=\pi+i\kappa$$ with: $$\omega^2=2(1+\cosh \kappa)$$ which is one to one for $$\omega>2$$ and $$\kappa>0$$. Now (up to an unimportant sign): $$A=\frac{1}{\sinh(\kappa(N+1))}$$ so there are no poles since $$\kappa>0$$. This second case to consider disappears in the continuum limit.

From the comment discussion, you would get similar results if you model your forcing by an inhomogeneous term in the bulk: $$\ddot y+2y_n-y_{n+1}-y_{n-1}=f_n$$ In this case, the analysis is even faster. Just decompose your force term in the eigenmode basis, so you get in the new basis uncoupled forced harmonic oscillators: $$\ddot Y_n+\omega_n^2Y_n=F_n$$ You there directly obtain the transfer function by looking at the frequency response and obtain a pole at each $$\omega_n$$.

Hope this helps.