In this video we're deriving a general solution for the simple pendulum so what we want to get
out of this is an expression for theta as a function of time so the angle as a function
of time for the pendulum once we have that we'll take note of what the period of the
oscillations is and then we have a simple example to apply that formula at the end
so i'm going to start by putting in the force of gravity on this thing and then we notice that this
is the angle theta as well and that allows us to decompose the force of gravity into components
the radial one being mg cosine theta and the tangential one mg
sine of theta now that tangential one is the one that we care about because that's what
causes a torque on this pendulum and causes it to accelerate in the angular sense so we're going to
apply the rotational equivalent of newton's second law that's tau equals i alpha and my angle is
shrinking here so this is a tricky point about the pendulum my angle is shrinking and that means the
angular acceleration must be a negative number in other words we need to count our torque as
negative so my torque is going to be negative sine theta that's the perpendicular component of force
multiplied by lever arm which is l and then i have to use the formula for the moment of inertia
of a point mass which is m r squared but our distance to the rotation axis in this picture is
l so ml squared times the angular acceleration alpha so alpha is the second time derivative
of the angle and i'm going to isolate that on the left hand side and write d squared theta
dt squared i'm going to cancel one factor of l while i'm at it and then the m also cancels
so dividing that l on both sides i get a negative g over l sine of theta
and now we're looking at a second order differential equation and the problem with this
differential equation is that it's non-linear so that sine of theta in there it makes it basically
impossible to solve this thing so the standard trick here is to expand the sine function in a
power series and i'll post a link to the video where i derive how to do this but as it turns
out the sine function can be expanded like this so theta minus theta cubed over three factorial
plus theta to the fifth over five factorial minus and so on and so on this is sometimes called the
maclaurin series and then the argument is if theta is pretty small then higher powers of theta are
going to be even smaller and therefore negligible and so we're allowed to truncate this series
and say that the sine of theta is approximately equal to theta
if theta is quote unquote small now how small does theta have to be for this to be a justifiable
approximation i ran some numbers today and i just wanted to look at 15 degrees or pi over 12
and the sine of pi over 12 is 0.259 and pi over 12 itself
so this is an important point this formula works if we're talking about theta in radians
pi over 12 itself is 0.262 and that comes out to about a 1 difference so with a 15 degree
oscillation we're talking about only a 1 error in our calculation and if you make theta smaller than
that it gets even more accurate okay replacing sine theta with this truncated power series
now we have a linear second order differential equation
and the solutions to these are guessable so i'll post a link to where i derived the solutions for
the simple harmonic oscillator because i explained in a lot of detail how you get the guessing done
but the bottom line is if i want a function that i differentiate twice to get back a minus sign and
then some constants and then the original function the general solution is given by sines and cosines
with a factor of root g over l in front of the t and i do have space here to verify that this
actually solves the differential equation i'm trying to show that if i differentiate twice
i get back a negative g over l in front of the original function so let's differentiate once
and i get negative square root g over l that's a result of the chain rule c1 sine
root g over lt plus root g over l c2 cosine root g over l t now i differentiate again
and the chain rule produces another factor of root g over l and when i multiply those
two factors i just get g over l so i have negative g over l c one cosine root
g over l t minus g over l c 2 sine
root g over l t and then i'm trying to find my original function in that
and if i factor out the negative g over l i do find it and if i express it as the original
theta of t that proves that this thing obeys the differential equation i was trying to solve so it
is a solution so there's our general solution for theta as a function of time for simple pendulum
and you can do problems where you're given initial conditions for the pendulum like an initial
angular position and angular velocity and then you could find the angle as a function of time for any
moment in time all i'm doing in this video is nailing down what's the period of oscillation so
these periodic functions the coefficient of t is root g over l so to get the period of oscillation
i just take 2 pi divided by that coefficient or 2 pi times the reciprocal of that coefficient
and there's our standard formula for the period of a simple pendulum applying this to a quick example
we're asked to get the period of oscillations for a simple pendulum of length 5 meters
so i just plug in 5 for l 9.8 for g and crunch the numbers and i get 4.49
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