Question

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

Answer 1

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 `bb((6))` is an oscillatory solution, with initial angle `vectheta_0`. This holds true for small angles of `theta`, such as less than `5^@`.

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.

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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 `mveca` and get:

`-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 `sintheta ~~ theta` for small angles of displacement. Therefore, taking the second derivative of this with respect to time gives:

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

Substituting `bb((2))` into `bb((1))`, we get:

`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, `e^(it) = cost + isint`, to write a linear combination of the assumed solution and rewrite it in terms of sines and cosines in the real domain:

`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 `c_1+c_2 = A` and `ic_1 - ic_2 = B` (arbitrary constants) to get:

`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 a starting angle `vectheta_0` (we set clockwise to be positive).
• `(dvectheta(t=0))/(dt) = 0`, i.e. the pendulum starts at rest and 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 `t = 0`, we have:

`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 `B = 0`, so from `bb((4))`, we get:

`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, `A = vectheta_0`, and we get the solution:

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

We recognize that `bb((6))` is an oscillatory solution, with initial angle `vectheta_0`. This holds true for small angles of `theta`, such as less than `5^@`.

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 `T` of the simple pendulum. Normally we write that the 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/(2pi) = f`, the frequency in `"s"^(-1)` (the `2pi` is the distance of 1 revolution in radians), and that `T = 1/f` is the period in `"s"`. Therefore:

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

Or, we have that:

`T = (2pi)/omega`

and therefore:

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