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

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

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Answer 1 · 2017-04-04T06:51:52

This is the solution we get at the end:

$\to θ (t) = {\to θ}_{0} cos ⎛ ⎜ ⎝ \sqrt{\frac{\to g}{L}} t ⎞ ⎟ ⎠$ $vectheta(t) = vectheta_0cos(sqrt(vecg/L)t)$ $" "$ $(6)$ $bb((6))$

$T = 2 π \sqrt{\frac{L}{\to g}}$ $T = 2pisqrt(L/(vecg))$ is the period, and $\to g > 0$ $vecg > 0$ .

We recognize that $(6)$ $bb((6))$ is an oscillatory solution, with initial angle ${\to θ}_{0}$ $vectheta_0$ . This holds true for small angles of $θ$ $theta$ , such as less than $5^{\circ}$ $5^@$ .

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.

DISCLAIMER: LONG ANSWER/DERIVATION!

The free-body diagram looks like this:

where $\to g > 0$ $vecg > 0$ 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 $m \to a$ $mveca$ and get:

$- m \to g sin \to θ = m \to a = m \frac{d^{2} \to x}{{d t}^{2}}$ $-mvecgsinvectheta = mveca = m(d^2vecx)/(dt^2)$

where $\frac{d^{2} \to x}{{d t}^{2}}$ $(d^2vecx)/(dt^2)$ is the second derivative of position with respect to time, and is the definition of acceleration.

Cancel out the mass to get:

$\frac{d^{2} \to x}{{d t}^{2}} = - \to g sin \to θ$ $(d^2vecx)/(dt^2) = -vecgsinvectheta$ $(1)$ $" "bb((1))$

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

$\to x = L sin \to θ$ $vecx = Lsinvectheta$

where $L$ $L$ is the length of the pendulum.

We take $sin θ \approx θ$ $sintheta ~~ theta$ for small angles of displacement. Therefore, taking the second derivative of this with respect to time gives:

$\frac{d^{2} \to x}{{d t}^{2}} \approx L \frac{d^{2} \to θ}{{d t}^{2}}$ $(d^2vecx)/(dt^2) ~~ L(d^2vectheta)/(dt^2)$ $(2)$ $" "bb((2))$

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

$L \frac{d^{2} \to θ}{{d t}^{2}} \approx - \to g \to θ$ $L(d^2vectheta)/(dt^2) ~~ -vecg vectheta$

$\frac{d^{2} \to θ}{{d t}^{2}} = - \frac{\to g}{L} \to θ$ $(d^2vectheta)/(dt^2) = -vecg/Lvectheta$

$\frac{d^{2} \to θ}{{d t}^{2}} + \frac{\to g}{L} \to θ = 0$ $color(green)((d^2vectheta)/(dt^2) + vecg/Lvectheta = 0)$ $(3)$ $" "bb((3))$

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 $(6)$ $bb((6))$ , the result.

If you do know differential equations, then you should read this.

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

$\to θ (t) = e^{r t}$ $vectheta(t) = e^(rt)$

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:

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

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

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