Math 160 Lab 2: Application

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Calculus 1 Lab 2
Linear Approximation Application

This section of the lab walk through an application using linear approximation.

Throughout this lab unless stated otherwise, compute all values to $6$ decimal places.

Let’s apply this notion of linear approximations in a physics application to answer a real-life question!

Consider a pendulum with ball $B$ with mass $m$ hanging from a string with length $L$ hanging from a block of woods at pivot point $P$ . Below is a picture of the pendulum that is displaced with angle $\theta$ from equilibrium.

In this model when the mass is directly below the pivot point $P$ (on the dashed line), the string is in equilibrium and doesn’t move.

Suppose we’d like to find out how long it takes this pendulum to swing back and forth (one period).

First of all, to find the period of the pendulum, we need to understand how the pendulum moves. Note that the position of the pendulum is completely determined by the angle of displacement $\theta$ , assuming that the string is always tight. If we move the ball of the pendulum to a certain $\theta \neq 0$ and let go, the pendulum will swing back and forth. Therefore the angle $\theta$ is a function of time, call it $t$ . So how does this pendulum move?

Newton’s second law of motion implies the following relationship holds in this system:

$-m g L \phantom {.} \sin (\theta ) = m L^2 \phantom {.}\dfrac {d^2 \theta }{d t^2}$ where $g$ is earth’s gravitational constant.

Our goal is to solve this differential equation to find what the function $\theta$ is. Once we find $\theta$ , we can find the period of the pendulum’s swing.

Let’s get to work. Manipulate the above equation so that ‘ $0$ ’ is on one side (the left side) and the $\dfrac {d^2 \theta }{d t^2}$ term is positive.

$0 = \answer [given]{m L^2} \dfrac {d^2 \theta }{d t^2} + \answer [given]{m g L \sin (\theta )}\\$

Hint: Remember that in Ximera, you type ‘ $\theta$ ’ by typing ‘theta’.

Note that $L,m,$ and $g$ are constants and are greater than $0$ , so we can divide the above equality by any of these terms to get an equivalent statement. Manipulate the above equality so that the coefficient of $\dfrac {d^2 \theta }{d t^2}$ is $1$ . Be sure to simplify fractions.

$0 = \dfrac {d^2 \theta }{d t^2} + \dfrac {\answer [given]{g}}{\answer [given]{L}} \phantom {.} \sin (\theta )$

Does the above relationship depend on the length of the string?

Yes No

There’s no way to eliminate $L$ from the equation, so the solution to the differential equation depends on the length of the string $L$ .

Does the above relationship depend on the gravitational constant?

Yes No

Same here. There’s no way to eliminate $g$ from the equation, so the solution to the differential equation depends on earth’s gravitational constant $g$ .

Does the above relationship depend on the mass of ball?

Yes No

By dividing by $m$ , we were able to eliminate the mass $m$ from the equation. Therefore the solution to the differential equation (the position/path of the ball) doesn’t depend on the mass of the ball. Even though $m$ was in the initial equality, it isn’t relevant to the solution. Holy smokes!

Now let’s try to solve this differential equation!

$0 = \dfrac {d^2 \theta }{d t^2} + \dfrac {g}{L} \sin (\theta )$

We will utilize the fact that $\theta$ , the angle of displacement of the ball $B$ ), is a function of time, $t$ . We are trying to solve the differential equation above so that we can find the period of the swing of the pendulum.

Manipulate the above equation to so that $\dfrac {d^2 \theta }{d t^2}$ is in terms of $\theta$ .

$\dfrac {d^2 \theta }{d t^2} = \answer [given]{-g\sin (\theta )/L}$

The differential equation is too tough for us to solve at the moment, but we can solve something similar.

Let’s solve this differential equation for small displacements of $\theta$ . We will replace ‘ $\sin (\theta )$ ’ with a linear approximation of $\sin (\theta )$ .

Question. If we want to solve this differential displacements from equilibrium, at value of $\theta$ should we approximate $\sin (\theta )$ ?

We find a linear approximation of $\sin (\theta )$ at $\theta = \answer [given]{0}$ .

Question. What is the linear approximation of $\sin (\theta )$ at the above $\theta$ value?

$\sin (\theta ) \approx \answer [given]{\theta }$

Hint: Remember $\theta$ is the variable, not $t$ nor $x$ . (Click the blue arrow to the right for another hint)

You’ve already found this early this earlier in this lab, at the end of the last section.

You found that sin(x) was approximately x near x=0.

$\phantom {.}$

Remark. The above approximation is actually a surprisingly good for $|\theta |\leq \frac {\pi }{12} = 15^\circ$ , so in this context, ”small” angles are angles that deviate no more that $15^\circ$ from equilibrium, by the way.

$\phantom {.}$

Back to the problem at hand, we are trying to find a solution to

$\dfrac {d^2 \theta }{d t^2} = -\dfrac {g}{L} \theta$

or if we rename the constant $C=\frac {g}{L}$ , we are trying to find a solution to

$\dfrac {d^2 \theta }{d t^2} = -C \theta .\\$

Another way to phrase this is that $\theta$ is a functions whose second derivative is $-1$ times a scalar $C$ of the original function.

Question. Do you know any types of function whose second derivative is $-1$ times itself (disregarding the $C$ )? Which of the following have at least one function that has this property.

Polynomials of degree $0$ (Constant functions) Polynomials of degree $1$ or bigger Trigonometric functions

Hint: Click the blue arrow to the right to reveal a hint.

If g(x)=0 (the horizontal line at height 0), what is its derivative?

Problem. Which polynomial $f(t)$ has the property that it’s second derivative is $-1$ times itself i.e. if $\theta =f(t)$ , then $\dfrac {d^2 \theta }{d t^2} = f''(t) = -f(t)$ ?

The polynomial function $f(t) = \answer [given]{0}$ has the property that $f''(t) = -f(t)$ .

$\phantom {.}$

Hint: Click the blue arrows on the right to reveal hints.

It is a constant function.

What is the derivative of a constant function?

The function is its own derivative.

What is the derivative of the zero function?

$\phantom {.}$

This is our equilibrium solution! This function has $\theta =0$ at $t=0$ . In other words, this is our solution when we put the ball directly below the point that the string is hanging from. This is no surprise though since this is always at equilibrium. What we really want is to find out what happens when we place the ball somewhere where the ball actually swings.

Question. Which of the following trigonometric functions have the property that $\dfrac {d^2 \theta }{d t^2} = -\theta$ ? Select all such functions.

$\sin (t)$ $\cos (t)$ $\tan (t)$ $\csc (t)$ $\sec (t)$ $\cot (t)$

$\phantom {.}$

Question . Since we are simulating the motion of the pendulum when letting go of the pendulum from a stand still, which of the functions you picked above would accurately simulate this situation?

$\sin (t)$ $\cos (t)$ $\tan (t)$ $\csc (t)$ $\sec (t)$ $\cot (t)$

Hint: Click the blue arrow to the right to reveal hints.

We want that the pendulum is not moving at t=0.

We want a function f(t) such that f’(t) = 0 at t=0.

We’re almost there! We know what the function generally looks like, but we need to check the fine details to make sure we answer the question we set out to answer.

We want to solve the differential equation

$\dfrac {d^2 \theta }{d t^2} = -\dfrac {g}{L} \theta$

or for the constant $C=\frac {g}{L}$ ,

$\dfrac {d^2 \theta }{d t^2} = -C\theta .$

For $\theta = \cos (at+b)$

$\begin{align*} \dfrac{d^2}{d t^2}(\theta) &= \dfrac{d^2}{d t^2}\Big(\cos(a t + b)\Big)\\ &= \dfrac{d}{d t} \Big( \dfrac{d}{dt}\Big(\cos(a t + b) \Big) \Big)\\ &= \dfrac{d}{d t} \Big( \answer[given]{- a \sin(a t+b)}\Big)\\ &= \answer[given]{-a^2 \cos(a t+b)}\\ &= -a^2 \theta \end{align*}$

Question. So if we are looking for a solution to $\dfrac {d^2 \theta }{d t^2} = -\dfrac {g}{L} \theta$ , then solutions to this differential equation is $\theta = \cos (at+b)$ where

$b = \answer [given]{0}$

and

$a^2 = \answer [given]{g/L}.$

Assuming $a>0$ , then

$a = \sqrt {\dfrac {\answer [given]{g}}{\answer [given]{L}}}.$

Therefore your solution to the differential equation is $\answer [given]{ \cos ((\sqrt {g}t)/\sqrt {L})}$

Question. Given your solution to the differential equation above, what is the period of a ball with mass $m$ on a string with length $L$ which is displaced at ”small” angles?

$(\answer [given]{2\pi })\sqrt {\Big (\answer [given]{L/g}\Big )}$

Hint: For $A\sin (at+b)$ or $A\cos (at+b)$ , the period of the function is $\dfrac {2\pi }{a}$ .

If the ball starts at an initial angle of displacement $\theta _0$ (where $|\theta _0|<\frac {\pi }{12}= 15^\circ$ ), we can describe the position of the ball at any given time by

$\theta (t) = \theta _0 \cos \Big (\sqrt {\frac {g}{L}} t\Big ).$ Pretty neat, huh?

Question. Given a ball with mass $0.6$ kilograms hanging from a string with length $7$ meters, what is the period of a pendulum swing when the swing doesn’t deviate more than $\frac {\pi }{12}$ radians from a free hang (where gravitational constant $9.8$ meters per second)?

The period is approximately $\answer [tolerance=0.01]{5.31026}.$

Use the calculator below to get your numeric answer. For example, if you type ‘2pi*4’, you will see the numeric answer.

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Calculus 1 Lab 2 Linear Approximation Application

Calculus 1 Lab 2
Linear Approximation Application