Differential equation: spring-mass system

[jonnburton]

I have been following the explanation of this topic in my book and there are a couple of things that I don't understand. I wondered if anyone could help me clear these things up.

The solution to the differential equation governing the spring-mass system is:

$\displaystyle u =e^{-\frac{t}{16}} \left(2cos\frac{\sqrt{255}}{16}t + \frac{2}{\sqrt{255}}sin\frac{\sqrt255}{16}t\right)$

The book says this is equal to $\displaystyle \frac{32}{\sqrt{255}}e^{-\frac{t}{16}}cos\left(\frac{\sqrt{255}}{16}t - \delta\right)$

Where $\displaystyle \delta = 0.625$

I don't really see how this can be, because the above equation appears to be of the form $\displaystyle cos\beta cos\alpha + sin\beta sin\alpha$, as opposed to $\displaystyle cos\beta cos\alpha - sin\beta sin\alpha$. So I would have said the solution reduces to:$\displaystyle \frac{32}{\sqrt{255}}e^{-\frac{t}{16}}cos\left(\frac{\sqrt{255}}{16}t + \delta\right)$



One other problem that I'm not sure how to approach is is to find $\displaystyle \tau$ such that $\displaystyle |u(t)|<0.1$ for all $\displaystyle \tau>t$

The only way I can see how to do this calculation is to say (using the book's equation from above):

$\displaystyle 0.1=\frac{32}{\sqrt255}e^{-\frac{t}{16}} cos\left(\frac {\sqrt255}{16}t -0.0625\right)$

$\displaystyle \frac{0.1\sqrt{255}}{32}=e^{-\frac{t}{16}}cos\left(\frac{\sqrt255}{16}-0.0625\right)$

However, from this I can see no way to isolate t.

 

[jonnburton]

I have just realised the answer to the first part of my post: I had copied down the identities for $\displaystyle cos(A \pm B)$ wrong... So I can see how the solution reduces to what the book says.

I still can't see how to solve for t ​though.

 

[Deleted member 4993]

I have been following the explanation of this topic in my book and there are a couple of things that I don't understand. I wondered if anyone could help me clear these things up.



One other problem that I'm not sure how to approach is is to find $\displaystyle \tau$ such that $\displaystyle |u(t)|<0.1$ for all $\displaystyle \tau>t$

The only way I can see how to do this calculation is to say (using the book's equation from above):

$\displaystyle 0.1=\frac{32}{\sqrt255}e^{-\frac{t}{16}} cos\left(\frac {\sqrt255}{16}t -0.0625\right)$

$\displaystyle \frac{0.1\sqrt{255}}{32}=e^{-\frac{t}{16}}cos\left(\frac{\sqrt255}{16}-0.0625\right)$

However, from this I can see no way to isolate t.

The best way to solve these problems is to approximate the answer using a graphical method then improve the accuracy of the answer by using a numerical method such as Newton's method (or use MS-Excel and use solver utility).

 

[jonnburton]

The best way to solve these problems is to approximate the answer using a graphical method then improve the accuracy of the answer by using a numerical method such as Newton's method (or use MS-Excel and use solver utility).

Thanks Subhotosh. So I was barking up the wrong tree trying to isolate the variable t. I'll have to look into the other methods.