Studying the r.m.s displacement of an oscillation by using Perseval's Theorem.

The r.m.s displacement, I am going to discuss about in this topic, is related to the displacement of an oscillation of a system under a periodic driving force. We can expand the periodic driving force by help of Fourier series and by using Perseval’s Theorem, we will check out its application in describing a driven damped oscillation.

Let’s consider an oscillation of a pendulum. It’s perturbed by a periodic driving force from its mean position and let deflect at an angle ϴ(theta). After each oscillation that force is acting on the system to diminish the effect of damping force. Now, let the bob of the pendulum goes x(t) distance after each perturbation of the driving force along the +ve X axis. Our goal is to calculate the r.m.s value of x(t) by using Perseval’s Theorem.

Solving the 2nd order differential equation, we get 

 x(t)= A cos(ωt -δ) which is the general equation of a

S.H.M where δ is the phase difference, A is amplitude and ω is the angular frequency of the oscillation.

If we now divide this displacement into nth parts then eqauation for each part becomes as below:

                   xn(t)=An cos(nωt- δn)

and therefore x(t) becomes as below:

x(t)= ∑ xn(t) = ∑ An cos(nωt- δn) where n=0 to ∞

An = fn/√(ω0²-n²ω²)²+4β² n²ω² is the amplitude of the

oscillation.

           δn=arctan(2βnω/ ω0²-n²ω²)

we get the representation for An and δn by doing complext exponential calculations or something like that. To show the whole calculation will make this more complicated. Here An is the amplitude of each nth term and δn is the corresponding phase difference. ω0 is the natural frequency of oscillation when there is no damping present which has the value 

                           ω0= √k/m

and ω is the angular frequency of the driving force.

Now we have to calculate the r.m.s value of x(t)

therefore, xr.m.s=√(<x²>)

Now, x2= ∑ ∑ Am cos(mωt- δm) An cos(nωt- δn)

therefore, <x2> =1/T* ∫x²dx ; limit runs from -T to T...............(1)

After calculation, we have the result;

∫cos(mωt- δm) *cos(nωt- δn) = T if m=n=0

                                                =T/2 if m=n

                                                =0 if m & n are different

in (1), only the terms with m=n should be retained and the result

becomes:

<x2>= A0² +1/2* ∑An² where n=1 to ∞

therefore,  xr.m.s=√(<x²>)

 

 or, xr.m.s=√( A0² +1/2* ∑An²)

Here, we have only considered the cosine terms and neglected the sine terms for easy calculation.  

Importance: This topic has many important uses but for our purposes, we will describe this; as we know to calculate An, δn by doing complex exponential calculations and using Perseval’s Theorem lets us know the responses <x2> for the oscillator. By this above mentioned topic, we can easily calculate the approximation for <x2>. Here is an example to explain the use of this theorem as an application of a driven damped oscillation.

The R.M.S displacement of a driven oscillator:

If we consider a simple example to understand the use of Fourier series in expanding the external periodic driving force acting on the pendulum, we can now do a very easy calculation.

Let’s consider a periodic force f(t) whose time period is T.

Now as this force is periodic, so this can be expressed as below:

f(t) =f(T+t) where T is the time period

According to Fourier Series,

f(t) = ∑{an cos(nωt) + bn sin( nωt)}....................(2)

or, f(t) = a0 + ∑an cos(nωt) where n=1 to ∞ ; 

we have considered only the cosine terms to simplify the

calculation, otherwise both sine and cosine terms have

significance in the calculation.

 From (2), we can derive that,

 

an=2/T*∫f(t) cos(nωt)dt

 

bn=2/T*∫f(t) sin(nωt)dt

where limit runs from (-T/2) to (+T/2) for n≥1

Let’s consider an oscillation under the effect of a square pulse(periodic driving force) having time period T. Our target is to plot a graph of this oscillation with xr.m.s vs T. Previously, we got to know 

            <x²>= A0²+ 1/2* ∑An²

if now this eqn. is truncated and expanded up to six terms and then find the corresponding <x2> and plot the graph xr.m.s vs T, we will notice some properties in this graphical representation. At each time, when T crosses an integer multiple of T0( natural time period of oscillation without any damping) that means T=nT0, we notice a sharp peak in the plot, as the nth Fourier component is at resonance. On the other hand, the next peak is lower than the previous one because the response of the oscillator decreases a little than its predecessor. As long as the driving force acts on the body, the resultant effect is wished to get diminished each time. So according to these properties, if we plot the provided data of xr.m.s and T with the help of any programming language, we will get a graph as shown below;