Differential Equation For The Pendulum (derivation)

The underlying condition of a mechanical framework (the totality of positions and speeds of its points at some moment of time) particularly decides the greater part of its movement. It is difficult to doubt this reality, since we learn it early. One can envision a world in which to determine the future of a system one should likewise know the acceleration at the initial moment, however experience demonstrates to us that our world is not like this. Numerous interesting ordinay differential equations (ODEs) emerge from applications. One purpose behind comprehending these applications in a mathematics class is that you can combine your physical intuition with your mathematical intuition in the same problem. Generally the result is an improvement of both. One such application is the motion of pendulum, i.e. a ball of mass m suspended from a perfect inflexible rod that is fixed at one end. The issue is to depict the movement of the mass point in a steady gravitational field.

Let

L = length of the rod measured, say, in meters,
 m = mass of the ball measured, say, in kilograms,
g = acceleration due to gravity = 9.8070 m/s2.

• The rod or cord on which the bob swings is massless, inextensible and always remains taut;
• The bob is a point mass;
• Motion occurs only in two dimensions, i.e. the bob does not trace an ellipse but an arc.
• The motion does not lose energy to friction or air resistance.
• The gravitational field is uniform.
• The support does not move.

Newton’s equations for the motion of a point x in a plane are vector equations

F = ma,

where F is the sum of the forces acting on the the point and a is the acceleration of the point, i.e.

a = d2x/dt2.

Since acceleration is a second derivative with respect to time t of the position vector, x, Newton’s equation is a second-order ODE for the position x. In x and y coordinates Newton’s equations become two equations

Fx = m d2x/dt2 ,                              Fy = m d2y/dt2 ,

where F_x and F_y are the x and y components, respectively, of the force F . From the figure (note definition of the angle θ) we see, upon resolving T into its x and y components, that

Fx = −T sinθ ,                             Fy = T cosθ − mg.

Substituting these expressions for the forces into Newton’s equations, we obtain the differential equations

X-> −T sinθ = m d2x/dt2 ,                Y-> T cosθ − mg = m d2y/dt2.

From the figure we see that

 A-> x = L sinθ,             B-> y = L − L cosθ.

The origin of the xy-plane is chosen so that at x = y = 0, the pendulum is at the bottom.
Differentiating A and B with respect to t, and then again, gives

x’ = L cosθ θ’,
x” = L cosθ θ” − L sinθ (θ’)2
y’ = L sinθ θ’,
y” = L sinθ θ” + L cosθ (θ’)2.

Substitute these in X and Y to obtain

−T sinθ = m L cosθ θ” − L sinθ (θ’)2,         T cosθ − mg = m L sinθ θ” + L cosθ (θ’)2.

Now multiply first equation by cos θ, and second by sin θ, and add the two resulting equations to obtain

−mg sinθ = mL θ”,

or

θ” + g/L sinθ = 0.