+251 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?
+251 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
+129847 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
+118667 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?
+26387 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:
+118667 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.
+33661 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!
.
+26387 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:
