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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
Sort:

#1
+121
+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
+79818
+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
+91226
+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
+18777
+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
edited by heureka  Mar 30, 2017
edited by heureka  Mar 30, 2017
#6
+91226
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
edited by Melody  Mar 30, 2017
#7
+26365
+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
edited by Alan  Mar 30, 2017
#10
+18777
+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
#8
+91226
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Thanks Alan, yes the results in W|A are weird.

I know you said more than that, but that is the bit i understand

Melody  Mar 30, 2017
#9
+26365
+1

Examples:

.

Alan  Mar 30, 2017
#12
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Can some please explain to me why t = 80? I get the read but why 80?

Guest Apr 5, 2017
#13
+26365
0

Think of the 80 as 2*40. Then the first set of results from Wolfram Alpha would look like:

40*2c1

40*(2c1 - 1/2)

40*(2c1 + 1/2)

etc.

where c1 is an integer.

.

Alan  Apr 6, 2017
#11
+91226
0

Thanks Heureka and Alan :))

Melody  Mar 30, 2017

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