Let x1(t) and x2 (t) be functions that represent the horizontal displacement of two simple pendulums from their central position at time t in seconds.

f both pendulums start at the same displacement at time t= 0, at what time

are they both next at the same displacement?

vest4R
Mar 30, 2017

#1**+1 **

No idea how to go about this.. I imagine you set them both equal to eachother and then solve for t... but how is the question lol

vest4R
Mar 30, 2017

#3**+2 **

Not sure how to solve something like this algebraically [but.....somebody else on here might know how!!!!]......thus.....I'll present a graphical solution :

It appears that they will be at the same displacement at about t = 5.268 seconds

CPhill
Mar 30, 2017

#4**+1 **

Thanks Chris,

hi vest4R,

I would do it exactly the same as Chris, just from a rough hand sketch it is easy to see that the answer lies between t=5 and t=6.

I do not know how to solve it algebraically.

I put this question into WolframAlpha and I am very confused by the information given.

The graph is the same as Chris and I produced and from their graph the answer is t = 5.268

https://www.wolframalpha.com/input/?i=0.5sin(pi*t%2F5)%3D0.1sin(pi*t%2F4)

BUT

The answer that they give is

\(t=40n\qquad n\in Z\\ and\\ t=20(2n+1) \qquad n \in Z \)

This makes NO sense to me what so ever!

**Maybe another mathematician can explain to me what WolframAlpha is talking about?**

Melody
Mar 30, 2017

#5**+3 **

Let x1(t) and x2 (t) be functions that represent the horizontal displacement of two simple pendulums from their central position at time t in seconds.

If both pendulums start at the same displacement at time t= 0,

at what time are they both next at the same displacement?

**WolframAlpha (Real Solutions) :**

\(\begin{array}{|l|rcll|} \hline x(t)=0 & t &=& 80\cdot c_1 \quad & | \quad c_1 \in \mathbb{Z} \\ & t &=& 40\cdot(2c_1-\frac12) \quad & | \quad c_1 \in \mathbb{Z} \\ & t &=& 40\cdot(2c_1+\frac12) \quad & | \quad c_1 \in \mathbb{Z} \\ & t &=& 40\cdot(2c_1+1) \quad & | \quad c_1 \in \mathbb{Z} \\ \hline \end{array}\)

\(\begin{array}{|l|rcll|} \hline x(t)\ne 0 & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 -2\cdot \arctan(\sqrt{0.1928427661680941676676430}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 -0.8275700335453974790842315\ldots ) \\\\ & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 +2\cdot \arctan(\sqrt{0.1928427661680941676676430}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 +0.8275700335453974790842315\ldots ) \\\\ & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 -2\cdot \arctan(\sqrt{1.102624388267996959021457}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 -1.6196234867202913859018615\ldots ) \\\\ & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 +2\cdot \arctan(\sqrt{1.102624388267996959021457}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 +1.6196234867202913859018615\ldots ) \\\\ & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 -2\cdot \arctan(\sqrt{6.348976379766929903229600}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 -2.3859824690979305397688904\ldots ) \\\\ & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 +2\cdot \arctan(\sqrt{6.348976379766929903229600}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 +2.3859824690979305397688904\ldots ) \\\\ \hline \end{array} \)

**WolframAlpha:**

heureka
Mar 30, 2017

#6**0 **

Thanks Heureka :/

I have questions :)

Clearly \(x_1(0)=x_2(0)=0\)

They are both multiples of sin(0) which equals 0 so they must be zero!

So why are you considering the possiblility where they do not equal zero???

And yes I have seen WolframAlphas solution but it makes no sense to me.

From the graph I can see that t = 5.258 is the first time the two pendulums are in the same place not including t=0

So where does t = any of those other numbers ??

I already have seen wolframAlpha results - I would like them to be explained to me.

**To me they make no sense.**

Melody
Mar 30, 2017

#7**+1 **

Melody,

The first group of Wolfram Alpha results ( e.g. 80.c1) give values that make x1 and x2 not only equal, but equal to zero, when c1 is an integer. The second group (e.g. 6.3662(6.28319.c1 - 0.82757 ) ) give values that make x1 and x2 equal (but not equal to zero) when c1 is an integer.

However, the Wolfram Alpha expressions don't seem to include *all *possible solutions as far as I can see. In particular they don't give the 5.268s result!

.

Alan
Mar 30, 2017

#10**+2 **

**However, the Wolfram Alpha expressions don't seem to include all possible solutions as far as I can see. **

**In particular they don't give the 5.268s result!**

Yes the Wolfram Alpha expressions gives all solutions also 5.268s.

Calculation of 5.268s with Wolfram Alpha:

Formula (Wolfram Alpha):

Let \(c_ 1 = 0\)

\(\begin{array}{|rcll|} \hline t &=& 6.3662(6.28319\cdot c_1 + 0.82757)\ \text{for} \ c_1 \in \mathbb{Z} \\ t &=& 6.3662(6.28319\cdot c_1 + 0.82757) \quad & | \quad c_1=0 \\ t &=& 6.3662(6.28319\cdot 0 + 0.82757) \\ t &=& 6.3662(0.82757) \\ t &=& 5.268\ldots \\ \hline \end{array}\)

see:

OK

or

\(\begin{array}{|rcll|} \hline t &=& 12.732(3.1416\cdot n+0.41379)\ \quad \ n \in \mathbb{Z} \\ t &=& 12.732(3.1416\cdot n + 0.41379) \quad & | \quad n=0 \\ t &=& 12.732(3.14169\cdot 0 + 0.41379) \\ t &=& 12.732(0.41379) \\ t &=& 5.268\ldots \\ \hline \end{array}\)

see:

heureka
Mar 30, 2017