Experimenting CLIL: school subjects in English / Maths / Physics / Science

The Simple Pendulum

Hi guys! Finally I’ve found some time to write here again.

Today I want to tell you something about the (mathematical) pendulum, and introduce it like you’ve never seen before at school.

You would say… why just the pendulum? Well, the pendulum is the simplest example of harmonic oscillator, and if you – like me – decide to study Physics, harmonic oscillators will be your bread and butter!

Furthermore, the equation of the motion of a pendulum is given “as it is” because you can’t get it with the simple mathematics of 3rd year at school. Don’t worry, I’m not going to use difficult mathematics, although it could be a bit difficult to understand if you aren’t a 5th year student.

Well, let’s start!

Imagine we have a simple pendulum made up of a thin string of length l. One end of the string is attached to the ceiling of our laboratory (on the surface of the Earth!), the other one has a heavy particle of mass m attached on it.

Task: we want to describe its motion with respect to the angle ϑ in the figure.

We start by setting two axes: one of these is tangent to the particle’s trajectory, and the other one is directed to the centre.                       We call them respectively →τ  and → n.

Now, let’s make some considerations about velocity and acceleration. The velocity tangential to the motion, let’s call it vτ, has module ddst, where ds is the infinitesimal arc travelled by the particle. In a circle, if the angle is given in radiant, we know that the arc is equal to the angle times the radius. So, our ds is equal to ldϑ, where is the infinitesimal angle corresponding to ds. Mathematically speaking, we have:

     ds   d(lϑ)   ldϑ    . vτ = ---= ----- = --- = lϑ      dt     dt     dt

where  . ϑ is the lazy physical notation to indicate the first derivative of ϑ with respect to the time t.

For the acceleration aτ, we have simply:

        .      .   .. aτ = d(lϑ-)=  ldϑ-=  lϑ       dt     dt

Again, .ϑ. indicates the second derivative of ϑ with respect to the time t.

Now, thanks to the IInd Law of Dynamics by Newton, we now that acceleration times mass in a given direction is equal to the sum of all the forces acting in that direction. Be careful: this is NOT a definition of force like some coursebooks say!

A force can be of various nature, and this law only tells us that a force (whatever it is) has a relationship with the acceleration in its direction. With this law, we can write:

  →    → m aτ = F

where → F is the force acting in the tangential direction to the motion. In this case, the force is equal to mg sin(ϑ). The sign is minus because the force is opposed to the direction of the motion. So, our equation is:

  ..                          ..    g //mlϑ = − //mg sin(ϑ)    = ⇒     ϑ = − --sin (ϑ )                                     l

For small angles, we use the approximation sin(ϑ) ϑ  (this approximation is given by the series expansion of the sine), obtaining:

..ϑ = − ω2ϑ

where we pose ω2 = g -l.

Now, we have to solve a quadratic differential equation…Aaargh!

Don’t worry, fortunately this is homogeneous and simply! To solve it, we assume that the solution is of the form ϑ = eλt for some λ. Substituting and deriving:

 2 λt   2 λt λ /e/ + ω /e/  = 0     =⇒      λ = ±iω

where i is the imaginary unit. At this point, to consider both solutions, we take their linear combination, obtaining:

ϑ = αeiωt + βe−iωt

for some α,β. Using Euler’s formula, the expression is equal to:

α (cos(ωt)+  isin(ωt))+ β (cos(ωt)−  isin(ωt)) =

= (α + β)cos(ωt)+ (α − β)isin(ωt )

Let us call (α + β) = c1 and (α β) = c2. We get:

c1cos(ωt)− c2isin(ωt) =
  ∘ -------(                                   ) =   c2 + c2   ∘--c1---cos(ωt)−  ∘--c2---isin(ωt)      1   2     c21 + c22           c21 + c22

Now, let φ be an angle so that:

( {cos(φ) = √-c1---             c21+c22 (sin(φ) = √-c22-2-             c1+c2

Using the addition formula of the cosine and calling ∘ -2---2-   c1 + c2 = A, we get:

ϑ(t) = A cos(ωt+ φ )

that is our solution.

The solution provided above is general, and A and φ are actually constants of integration. To find a particular solution, we need the so-called initial conditions. Let’s suppose, for example, that our pendulum is started with a null velocity and an initial angle, ϑ0. The initial conditions become:

{ .  ϑ (0) = 0  ϑ (0) = ϑ0

Deriving and solving the system, we get:

{   φ = 0   A = ϑ0

So, our particular solution is:

ϑ (t) = ϑ0cos(ωt)

As an exercise, we can imagine different initial conditions and find out the different particular solutions of the equation of our system. I leave this to your imagination! 😉

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