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December 11th, 2016, 12:43 PM   #1
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Question Differential equations on pendulum

Hi guys my professor is starting diff equations with us and gave us this question to look up online and try to see how it works, no matter where i look i cannot find the solution and i have never done these kind of diff equations before, only basic ones about population growth and stuff. They are usually e^kt, so very basic. Here is the question its in the attachment.


Took me ages to type properly haha. Anyways I would appreciate the answer as I have no clue and I just need to try and understand it. As you sol it could you please explain why you are doing what you are doing XD. Thanks so much guys. I had to register on this forum just to try and get the solution.
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December 11th, 2016, 01:18 PM   #2
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Originally Posted by greally2014 View Post
Hi guys my professor is starting diff equations with us and gave us this question to look up online and try to see how it works, no matter where i look i cannot find the solution and i have never done these kind of diff equations before, only basic ones about population growth and stuff. They are usually e^kt, so very basic. Here is the question its in the attachment.


Took me ages to type properly haha. Anyways I would appreciate the answer as I have no clue and I just need to try and understand it. As you sol it could you please explain why you are doing what you are doing XD. Thanks so much guys. I had to register on this forum just to try and get the solution.
Note that you don't actually have to solve the equation. You were given the solution for $\displaystyle \theta (t)$. So take the derivatives:
The equation for $\displaystyle \theta (t) $ should be $\displaystyle \theta = \theta _0~cos \left ( \sqrt{ \frac{g}{L}} t \right )$.

$\displaystyle \theta = \theta _0 ~ cos \left ( \sqrt{\frac{g}{L}} t \right )$

$\displaystyle \frac{d \theta }{dt} = -\theta _ 0 ~ \sqrt{\frac{g}{t}}~sin \left ( \sqrt{\frac{g}{L}} t \right )$

$\displaystyle \frac{d^2 \theta }{dt^2} = -\theta _0 \frac{g}{t}~ cos \left ( \sqrt{ \frac{g}{L}} t \right )$

Now plug that into your differential equation and see if it works.

-Dan

Addendum: You should also check that the given $\displaystyle \theta (t)$ matches the initial conditions.

Last edited by topsquark; December 11th, 2016 at 01:24 PM.
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December 11th, 2016, 01:45 PM   #3
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I think i get it...

So if i understand, you know that θ(0) = θo

and you know what θ(t) is for all t.

in other words, θ = θo, and we have θ(t)?

So we want an expression for θ using both. So why do you multiply θ(t) by θo?

the rest is basic differentiation

But when you are deriving sin and cos, you are multiplying by the derivative inside that brackets of sin and cos which i get, but how is the derivative of the root of the gravity over length multiplied by t = the root of gravity over t?

and what do you mean when you say plug into diff equation, i have never done these before my job is to try and understand like my teacher wants, we arent expected to be able to do this
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December 11th, 2016, 02:05 PM   #4
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I think i get it...

Quote:
Originally Posted by topsquark View Post
Note that you don't actually have to solve the equation. You were given the solution for $\displaystyle \theta (t)$. So take the derivatives:
The equation for $\displaystyle \theta (t) $ should be $\displaystyle \theta = \theta _0~cos \left ( \sqrt{ \frac{g}{L}} t \right )$.

$\displaystyle \theta = \theta _0 ~ cos \left ( \sqrt{\frac{g}{L}} t \right )$

$\displaystyle \frac{d \theta }{dt} = -\theta _ 0 ~ \sqrt{\frac{g}{t}}~sin \left ( \sqrt{\frac{g}{L}} t \right )$

$\displaystyle \frac{d^2 \theta }{dt^2} = -\theta _0 \frac{g}{t}~ cos \left ( \sqrt{ \frac{g}{L}} t \right )$

Now plug that into your differential equation and see if it works.

-Dan

Addendum: You should also check that the given $\displaystyle \theta (t)$ matches the initial conditions.
So if i understand, you know that θ(0) = θo

and you know what θ(t) is for all t.

in other words, θ = θo, and we have θ(t)?

So we want an expression for θ using both. So why do you multiply θ(t) by θo?

the rest is basic differentiation

But when you are deriving sin and cos, you are multiplying by the derivative inside that brackets of sin and cos which i get, but how is the derivative of the root of the gravity over length multiplied by t = the root of gravity over t?

and what do you mean when you say plug into diff equation, i have never done these before my job is to try and understand like my teacher wants, we arent expected to be able to do this
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December 11th, 2016, 02:16 PM   #5
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Originally Posted by greally2014 View Post
So if i understand, you know that θ(0) = θo

and you know what θ(t) is for all t.

in other words, θ = θo, and we have θ(t)?

So we want an expression for θ using both. So why do you multiply θ(t) by θo?

the rest is basic differentiation

But when you are deriving sin and cos, you are multiplying by the derivative inside that brackets of sin and cos which i get, but how is the derivative of the root of the gravity over length multiplied by t = the root of gravity over t?

and what do you mean when you say plug into diff equation, i have never done these before my job is to try and understand like my teacher wants, we arent expected to be able to do this
You are given a solution and the defining equation. All you need to do is calculate the derivatives and use them in the equation. In this case:
$\displaystyle \frac{d^2 \theta }{dt^2} + \frac{g}{L} \theta = 0$

so you need to verify that
$\displaystyle \left ( -\theta _0 ~ \frac{g}{L} ~ cos \left ( \sqrt{ \frac{g}{L}}~t \right ) \right ) + \frac{g}{L} \left ( \theta _0 ~ cos \left ( \sqrt{\frac{g}{L}} ~ t \right ) \right ) = 0$

Since you are starting off and the question gives you $\displaystyle \theta (t)$ then you don't need to know how to solve it. Your instructor will get there. But since you are curious:
$\displaystyle \frac{d^2 \theta }{dt^2} + \frac{g}{L} \theta = 0$

This is a second order linear homogeneous differential equation with constant coefficients. One method for solution uses the characteristic equation:
$\displaystyle m^2 + \frac{g}{L} = 0$

(Use a factor of m for each $\displaystyle \frac{d^m \theta }{dt^m}$)

The solution for m is $\displaystyle i \sqrt{ \frac{g}{L} }$. We have two forms we can use for this solution:
Either
$\displaystyle \theta = A e^{i \omega t} + B e^{-i \omega t}$

or
$\displaystyle \theta = A ~ cos( \omega t) + B ~ sin(\omega t)$

where $\displaystyle \omega = \sqrt{\frac{g}{L}}$ in each case. Both forms give the same result but since your question uses cosine we'll use the second form.

Now for the initial conditions. We know that $\displaystyle \theta (0) = \theta _0$ so
$\displaystyle \theta (0) = A~cos( \omega \cdot 0 ) + B~sin( \omega \cdot 0) = \theta _0$

$\displaystyle A \cdot 1 + B \cdot 0 = \theta _0$

Thus $\displaystyle A = \theta _0$.

To get B we use $\displaystyle \frac{d \theta }{dt} = -\theta _0 \omega ~ sin( \omega t) + B \omega ~cos( \omega t)$
where I've substituted $\displaystyle A = \theta _0$.

So
$\displaystyle \frac{d \theta}{dt} (0) = -\theta _0 \omega ~ sin(\omega \cdot 0) + B \omega ~cos( \omega \cdot 0) = 0$

$\displaystyle -\theta _0 \omega \cdot 0 + B \omega \cdot 1 = 0$

or B = 0.

Thus the solution is $\displaystyle \theta (t) = \theta _0 ~ cos( \omega t)$

It looks complicated but you'll get plenty of experience with these in your class. It's not so bad once you've done it a couple of times.

-Dan

Last edited by topsquark; December 11th, 2016 at 02:27 PM.
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December 11th, 2016, 02:26 PM   #6
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Quote:
Originally Posted by topsquark View Post
You are given a solution and the defining equation. All you need to do is calculate the derivatives and use them in the equation. In this case:
$\displaystyle \frac{d^2 \theta }{dt^2} + \frac{g}{L} \theta = 0$

so you need to verify that
$\displaystyle \left ( -\theta _0 ~ \frac{g}{L} ~ cos \left ( \sqrt{ \frac{g}{L}}~t \right ) \right ) + \frac{g}{L} \left ( \theta _0 ~ cos \left ( \sqrt{\frac{g}{L}} ~ t \right ) \right ) = 0$

Since you are starting off and the question gives you $\displaystyle \theta (t)$ then you don't need to know how to solve it. Your instructor will get there. But since you are curious:
$\displaystyle \frac{d^2 \theta }{dt^2} + \frac{g}{L} \theta = 0$

This is a second order linear homogeneous differential equation with constant coefficients. One method for solution uses the characteristic equation:
$\displaystyle m^2 + \frac{g}{L} = 0$

(Sort of use a factor of m for each $\displaystyle \frac{d \theta}{dt}$.)

The solution for m is $\displaystyle i \sqrt{ \frac{g}{L} }$. We have two forms we can use for this solution:
Either
$\displaystyle \theta = A e^{i \omega t} + B e^{-i \omega t}$

or
$\displaystyle \theta = A ~ cos( \omega t) + B ~ sin(\omega t)$

where $\displaystyle \omega = \sqrt{\frac{g}{L}}$ in each case. Both forms give the same result but since your question uses cosine we'll use the second form.

Now for the initial conditions. We know that $\displaystyle \theta (0) = \theta _0$ so
$\displaystyle \theta (0) = A~cos( \omega \cdot 0 ) + B~sin( \omega \cdot 0) = \theta _0$

$\displaystyle A \cdot 1 + B \cdot 0 = \theta _0$

Thus $\displaystyle A = \theta _0$.

To get B we use $\displaystyle \frac{d \theta }{dt} = -\theta _0 \omega ~ sin( \omega t) + B \omega ~cos( \omega t)$
where I've substituted $\displaystyle A = \theta _0$.

So
$\displaystyle \frac{d \theta}{dt} (0) = -\theta _0 \omega ~ sin(\omega \cdot 0) + B \omega ~cos( \omega \cdot 0) = 0$

$\displaystyle -\theta _0 \omega \cdot 0 + B \omega \cdot 1 = 0$

or B = 0.

Thus the solution is $\displaystyle \theta (t) = \theta _0 ~ cos( \omega t)$

It looks complicated but you'll get plenty of experience with these in your class. It's not so bad once you've done it a couple of times.

-Dan
Thank you, yes in your original answer you differentiated wrong and i got confused but you corrected it second time, i was confused, one i verify it is = 0, how do i calculate when it reaches vertical and max height? etc, what would theta be
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December 11th, 2016, 02:33 PM   #7
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Originally Posted by greally2014 View Post
Thank you, yes in your original answer you differentiated wrong and i got confused but you corrected it second time, i was confused, one i verify it is = 0, how do i calculate when it reaches vertical and max height? etc, what would theta be
Yes, I wound up doing a number of edits. Sorry for the confusion.

By "vertical" I presume you mean when the pendulum is at its lowest level (ie. the string is straight up and down)? That would correspond to $\displaystyle \theta = 0$. So let $\displaystyle \theta (t) = 0$ and solve for t.

Max height is the largest value of $\displaystyle \theta$. So you are finding the (absolute) maximum of $\displaystyle \theta$, which we know to be $\displaystyle \theta _0$. Find the first derivative and set it equal to 0, then solve for t.

-Dan
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December 11th, 2016, 03:08 PM   #8
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Quote:
Originally Posted by topsquark View Post
Yes, I wound up doing a number of edits. Sorry for the confusion.

By "vertical" I presume you mean when the pendulum is at its lowest level (ie. the string is straight up and down)? That would correspond to $\displaystyle \theta = 0$. So let $\displaystyle \theta (t) = 0$ and solve for t.

Max height is the largest value of $\displaystyle \theta$. So you are finding the (absolute) maximum of $\displaystyle \theta$, which we know to be $\displaystyle \theta _0$. Find the first derivative and set it equal to 0, then solve for t.

-Dan
Yes it just asks when does the pendulum first return to the vertical and does the time taken depend on the initial displacement.

All im doing is deriving what im given? and subbing it in to make sure its correct? after that im just using theta (t)? ok that seems cool
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December 11th, 2016, 03:14 PM   #9
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Quote:
Originally Posted by topsquark View Post
Yes, I wound up doing a number of edits. Sorry for the confusion.

By "vertical" I presume you mean when the pendulum is at its lowest level (ie. the string is straight up and down)? That would correspond to $\displaystyle \theta = 0$. So let $\displaystyle \theta (t) = 0$ and solve for t.

Max height is the largest value of $\displaystyle \theta$. So you are finding the (absolute) maximum of $\displaystyle \theta$, which we know to be $\displaystyle \theta _0$. Find the first derivative and set it equal to 0, then solve for t.

-Dan
Quote:
Originally Posted by greally2014 View Post
Yes it just asks when does the pendulum first return to the vertical and does the time taken depend on the initial displacement.

All im doing is deriving what im given? and subbing it in to make sure its correct? after that im just using theta (t)? ok that seems cool
The only thing i dont understand is why you multiply theta(0) by theta(t), thats the only thing xd
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December 11th, 2016, 03:42 PM   #10
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Originally Posted by greally2014 View Post
The only thing i dont understand is why you multiply theta(0) by theta(t), thats the only thing xd
Given
$\displaystyle \theta (t) = A~cos( \omega t) + B~sin( \omega t)$

From the initial condition we have $\displaystyle \theta (t = 0) = \theta _0$

$\displaystyle \theta _0$ is just a constant which defines what $\displaystyle \theta$ is at t = 0. So in the general solution we have:

$\displaystyle \theta (t = 0) = A~cos( \omega \cdot 0) + B~sin( \omega \cdot 0)$

$\displaystyle \theta (t = 0) = \theta _ 0 = A~cos( \omega \cdot 0) + B~sin( \omega \cdot 0)$

$\displaystyle \theta _ 0 = A \cdot 1 + B \cdot 0$

$\displaystyle \theta _ 0 = A$

-Dan
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