Free oscillations of a pendulum

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Christian Lawson-Perfect 3 years, 5 months ago

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Christian Lawson-Perfect 3 years, 5 months ago

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Question statement

In the absence of air resistance, the equation of motion of a bob, of mass $m$ , at the end of a pendulum of length $l$ is

$ml\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} = -mg\sin\theta\text{.}$

In the small angle approximation, $\sin\theta\approx\theta$ , this reduces to

$\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}+\frac{g}{l}\theta = 0\text{.}$

Give any introductory information the student needs.

			Name	Type	Generated Value

k	integer	3
beta	rational	3/8
damping5	decimal	dec("7.988619396143976")
damping0p01	decimal	dec("10.6666666666666656")
is_under_damped	boolean	true

			Name	Type	Generated Value

g	number	9.81
l	integer	4
m	integer	4
T	number	4.0121333614
theta0_frac	integer	3
theta0	number	1.0471975512

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..5)

JME function reference

Description

friction coefficient (kgs-1)

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

beta
is_under_damped

Advice
"Part d)" - prompt
"Part d)" → "Unnamed gap" - Minimum accepted value
"Part d)" → "Unnamed gap" - Maximum accepted value

Parts

Part a) Gap-fill
1. Gap Mathematical expression
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Part b) Gap-fill
1. Gap Number entry
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Part c) Gap-fill
1. Gap Number entry
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Part d) Gap-fill
1. Gap Number entry
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Part e) Gap-fill
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

{jsx_pendulum()}

You are told that $g=\var{g}\textrm{ms}^{-2}$ and that the length of the pendulum's rod $l = \var{l}$ m.

Solve for $\theta(t)$ with initial conditions $\frac{\mathrm{d}\theta}{\mathrm{d}t}(0)=0$, $\theta(0)=\frac{\pi}{\var{theta0_frac}}$.

$\theta(t) = $ Unnamed gap

Advice

a)

We have a second order constant coefficient differential equation,

\[ \frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}+\frac{g}{l}\theta = 0 \text{.} \]

Setting $\theta = e^{\lambda t}$, we obtain the auxiliary equation

\[ \lambda^2 + \frac{g}{l} = 0 \implies \lambda = \pm i\sqrt{\frac{g}{l}} \text{.} \]

The general solution for an auxiliary equation with complex roots is

\[ \theta(t) = A\sin\omega t + B\cos\omega t \text{,} \]

where $\omega = \sqrt{\frac{g}{l}}$ is the angular frequency of the oscillations, also known as the natural frequency: the frequency with which the pendulum oscillates in the absence of external forcing).

Note that

\[ \frac{\mathrm{d}\theta}{\mathrm{d}t} = A\omega\cos\omega t - B\omega\sin\omega t \text{.} \]

Applying the initial conditions:

\begin{align}
\frac{\mathrm{d}\theta}{\mathrm{d}t} (0) = 0 &=A\omega\cos(0) - B\omega\sin(0) \text{,} \\
&= A\omega \text{,} \\
\therefore A &= 0 \text{.}
\end{align}

\begin{align}
\theta (0) =\frac{\pi}{\var{theta0_frac}} &=B\cos(0) \text{,} \\
\therefore B &=\frac{\pi}{\var{theta0_frac}} \text{.}
\end{align}

Therefore our solution is

\[ \theta(t) = \frac{\pi}{\var{theta0_frac}}\cos\left(\sqrt{\frac{g}{l}}t\right)\text{.} \]

Using the given values for $g$ and $l$,

\[ \theta(t) = \frac{\pi}{\var{theta0_frac}}\cos\left(\sqrt{\frac{\var{g}}{\var{l}}}t\right)\text{.} \]

b)

The time period of oscillations is given by

\[ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{l}{g}} =2\pi\sqrt{\frac{\var{l}}{\var{g}}} = \var{precround(T,2)} \text{ s.} \]

c)

Differentiating our solution from part (a), we have

\[ \frac{d\theta}{dt} = -\frac{\pi}{\var{theta0_frac}}\omega \sin\omega t\text{.} \]

This is at its maximum when $\sin \omega t = -1$,

\[ \left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)_{\text{max}} = \frac{\pi}{\var{theta0_frac}}\omega =\frac{\pi}{\var{theta0_frac}}\sqrt{\frac{\var{g}}{\var{l}}} = \var{precround(theta0*sqrt(g/l),2)}\text{ rad s}^{-1 }\text{.} \]

d)

With the addition of a damping term, our differential equation becomes

\[ \frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}+\frac{k}{m}\frac{\mathrm{d}\theta}{\mathrm{d}t}+\frac{g}{l}\theta = 0\text{.} \]

This is still a constant-coefficient second-order differential equation, however the extra term changes the auxiliary equation. For convenience we can let $\beta = \frac{k}{2m}$ (we will later find that this is $\beta$ as asked for in the question) and $\omega^2 = \frac{g}{l}$, as before, to put our auxiliary equation in the form

\[ \lambda^2 + 2\beta \lambda + \omega^2 = 0\text{,} \]

which has the solutions

\[ \begin{align} \lambda &=-\frac{2\beta}{2}\pm \frac{\sqrt{2\beta^2-4\omega^2}}{2} \\
&= -\beta\pm \sqrt{\beta^2-\omega^2} \\
&= -\beta\pm i\sqrt{\omega^2-\beta^2}\text{.} \end{align}\]

Note that the values in our problem have determined that $\omega >\beta$. As in part (a), we have an auxiliary equation with complex roots:

\[\theta(t) = e^{-\beta t}\left[A\sin\left(\sqrt{\omega^2-\beta^2}\; t\right) + B\cos\left(\sqrt{\omega^2-\beta^2}\; t\right)\right]\text{.} \]

Using the values and initial conditions from part (a), this becomes

\[\theta(t) = \frac{\pi}{\var{theta0_frac}}e^{-\beta t}\cos\left(\sqrt{\frac{\var{g}}{\var{l}}-\beta^2}\; t\right)\text{,} \]

with

\[ \beta = \frac{k}{2m} = \frac{\var{k}}{2\times\var{m}} = \var{precround(beta,2)}\text{.}\]

e)

The amplitude of the oscillation is at 5% of the initial release angle when

\[ e^{-\beta t} = 0.05\text{.} \]

The time taken is therefore

\[ t = -\frac{\ln{(0.05)}}{\beta}=-\frac{\ln{(0.05)}}{\var{precround(beta,2)}} = \var{precround(damping5,2)}\text{ s.} \]

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Christian Lawson-Perfect 3 years, 5 months ago

Christian Lawson-Perfect 3 years, 5 months ago

Error in variable testing condition

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Free oscillations of a pendulum

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Christian Lawson-Perfect 3 years, 5 months ago

Christian Lawson-Perfect 3 years, 5 months ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

a)

b)

c)

d)

e)

Generated value: `integer`