# Show that the oscillation of a pendulum is a simple harmonic?

##### 1 Answer

This is the solution we get at the end:

#vectheta(t) = vectheta_0cos(sqrt(vecg/L)t)# #" "# #bb((6))#

#T = 2pisqrt(L/(vecg))# is the period, and#vecg > 0# .

We recognize that *oscillatory* solution, with initial angle ** small angles** of

*This is the restriction we place onto any simple harmonic oscillator; it works best for small angles of displacement if it's a pendulum, or for small horizontal cartesian displacement if it's a ball and spring.*

**DISCLAIMER:** *LONG ANSWER/DERIVATION!*

The free-body diagram looks like this:

where

#vecg > 0# and horizontally-outwards and down are positive (thus, horizontally-inwards and up are negative).

Since this is the main driving force, we set it equal to

#-mvecgsinvectheta = mveca = m(d^2vecx)/(dt^2)# where

#(d^2vecx)/(dt^2)# is the second derivative of position with respect to time, and is the definition of acceleration.

Cancel out the mass to get:

#(d^2vecx)/(dt^2) = -vecgsinvectheta# #" "bb((1))#

Now, we need to figure out what the acceleration is. We can start by knowing that the horizontal displacement is:

#vecx = Lsinvectheta# where

#L# is the length of the pendulum.

We take

#(d^2vecx)/(dt^2) ~~ L(d^2vectheta)/(dt^2)# #" "bb((2))#

Substituting

#L(d^2vectheta)/(dt^2) ~~ -vecg vectheta#

#(d^2vectheta)/(dt^2) = -vecg/Lvectheta#

#color(green)((d^2vectheta)/(dt^2) + vecg/Lvectheta = 0)# #" "bb((3))#

This is the differential equation for a simple oscillator. This is called a **linear second-order ordinary differential equation**.

If you do not know differential equations, you can skip all the way to equation#bb((6))# , the result.If you do know differential equations, then you should read this.

For this, we can assume a solution of the form:

#vectheta(t) = e^(rt)#

and we find:

#r^2cancel(e^(rt))^(ne0) + vecg/Lcancel(e^(rt))^(ne0) = 0#

#r^2 + vecg/L = 0#

#r = pmisqrt(vecg/L)#

We can then use Euler's formula,

#vectheta(t) = c_1e^(isqrt(vecg/L)t) + c_2e^(-isqrt(vecg/L)t)#

#= c_1[cos(sqrt(vecg/L)t) + isin(sqrt(vecg/L)t)] + c_2[cos(sqrt(vecg/L)t)-isin(sqrt(vecg/L)t)]#

#= (c_1+c_2)cos(sqrt(vecg/L)t) + (ic_1-ic_2)sin(sqrt(vecg/L)t)#

For convenience, call

#vectheta(t) = Acos(sqrt(vecg/L)t) + Bsin(sqrt(vecg/L)t)# #" "bb((4))#

The pendulum has certain **initial conditions**:

#vectheta(t=0) = vectheta_0# , i.e. there is astarting angle#vectheta_0# (we set clockwise to be positive).#(dvectheta(t=0))/(dt) = 0# , i.e. the pendulumstarts at restand thus does not have a rate of change for the angle yet at time zero.

Apply the *second initial condition* by taking the first derivative, and then setting the time equal to zero:

#(dvectheta)/(dt) = -Asqrt(vecg/L)sin(sqrt(vecg/L)t) + Bsqrt(vecg/L)cos(sqrt(vecg/L)t)#

At

#cancel(-Asqrt(vecg/L)sin(sqrt(vecg/L)*0))^(0) + Bsqrt(vecg/L)cancel(cos(sqrt(vecg/L)*0))^(1) = 0#

The only way for this condition to hold true is if

#vectheta(t) = Acos(sqrt(vecg/L)t)# #" "bb((5))#

From the *first initial condition*, we have:

#vectheta(0) = vectheta_0 = Acancel(cos(sqrt(vecg/L)*0))^(1)#

Therefore,

#color(blue)(vectheta(t) = vectheta_0cos(sqrt(vecg/L)t))# #" "bb((6))#

We recognize that *oscillatory* solution, with initial angle ** small angles** of

*This is the restriction we place onto any simple harmonic oscillator; it works best for small angles of displacement if it's a pendulum, or for small horizontal cartesian displacement if it's a ball and spring.*

Lastly, we need to find the **period** *simple harmonic solution* is of the form:

#vectheta(t) = Acos(omegat)# ,where

#omega# is the angular frequency in#"rad/s"# .

In this case, we must have that:

#omega = sqrt(vecg/L)#

We also have the relationship that

#omega = 2pif = (2pi)/T#

Or, we have that:

#T = (2pi)/omega#

and therefore:

#color(blue)(T = 2pisqrt(L/(vecg)))#