math problem (sinusoidal motion)

[scully44]

Does anyone know how to solve the following equation:

-A*2*PI/6*COS(2*PI*t/6)-B*2*PI/18*COS(2*PI*t/18)=0

Where: A is a constant
B is a constant
t = time

Find: all values of t between 0 & 18 such that the above statment is true.

the function is velocity of a point subjected to sinusoidal motion of 2 frequencies & 2 amplitudes. Times of Max. acceleration to design some hydraulic actuators to impart said motion.

 

[swearingen]

The answer is that it depends on A and B. It's great that you say A and B are constant, but their values must be known before you solve this one. There are no general t values that will work for all selections of A and B.



If you "heard" it on the internet, it's guilty until proven innocent. - DCS

 

[handleman]

Ever heard of algebra? Trig identities?

(PI/3)*(-Acos(PI*t/3) - B/3*cos(PI*t/9))=0

PI/3 term drops out. Let x=PI*t/9

-Acos(3x)-B/3cos(x)=0

use triple angle trig identity:

-4A(cos(x)^3)+3Acos(x)-B/3cos(x)=0

let y = cos(x)

-4Ay^3+(3A-B/3)y=0

Problem is reduced to a simple cubic.

 

[rb1957]

from there it's even simpler ...
factor ...
y*(-4y^2+(3A-B/3)) = 0

y = 0, sqrt(3A-B/3)/2 (+ve and -ve)

and y = cos(pi/9*t) ...

 

[handleman]

Whoops, good catch, rb.

 

[eromlignod]

That reduces it, guys, but you still have to know A and B to solve for t, which is apparently what he wants.

You can find at least *some* of the values of t without knowing A and B though, since the equation is looking for zero-crossings.

If both the cosines are zero, then the equation is satisfied regardless of A and B, since they are multiplied by 0. Cosines are zero at pi/2, 3*pi/2, 5*pi/2, etc. If you set the terms inside the cosines to n*pi/2, where n is an odd integer, then pi*t/3 solves to

t = 1.5 n

and pi*t/9 solves to

t = 4.5 n

where n is an odd integer.

The only time t<18 that both these equations are true is when

t = 4.5

This may not be the only solution, but it is a solution regardless of the values of A and B.

Don
Kansas City

 

[rb1957]

i suspect that an "answer" in terms of A and B would meet the needs of the "problem"

 

[handleman]

Don, the complete analytical solution as completed by rb1957 yields your answer in the "y=0" solution that is independent of A and B, along with the other two answers in terms of A and B. It is implicitly obvious that to find actual values for the other two solutions one must know both A and B. However, the way the original equation was posted leaves me with doubts as to whether the OP is capable of following along and understanding the solution anyway.

 

[prost]

my two cents: assuming 't' is real valued, then cos(pi*t/9) is real valued. If so, then Sqrt(3A-B/3) has to be real, and this is true if and only if the argument in the Sqrt is positive or zero, that is, 3A>=B/3, or A>=B/9 (that is, A is greater than or equal to B/9). This tells you not what A and B are, only that A has a constraint, it must be greater than or equal to one-ninth of B.

 

[wes616]

Hi guys,

Not to be a pill... but why are we answering something that is obviously a homework problem?

Wes C.
------------------------------
No trees were killed in the sending of this message, but a large number of electrons were terribly inconvenienced.

 

[scully44]

Its not a homework problem and I am perfectly capable of following the solution along. I don't deal with these sort of equations on a regular basis as 99% of my work involves statics. I was hoping for a little help to save me the time of digging through text books I haven't looked at in 10 years. I know the values of A and B & posted the equation that way in hopes of getting an answer in terms of A and B so that they could be put in a spread sheet & get new values of t by changing the inputs. Yes it is obvious that t = 4.5 is one solution & t = 13.5 is another. I am interested in the other times when the cos of A cos(PI*t/6)=-B cos(pi*t/18). If A=B t=2.25,4.5 6.75 etc. IF B>A the only solution is t = 4.5 & 13.5 . But if A>B there will be 6 solutions t=4.5, t=13.5 & 4 others. The 4 others are the ones I'm interersted in. My lack of familiarity with trig identity functions prevented me from solving this equation in a timely manner.

Thanks for the help.

I'll chock up the condescending comments to not putting enough detail in my question.

 

[prex]

rb1957 made a little mistake above, and it turns out that cos([&pi;]t/9)=[&plusmn;][&radic;](0.75-B/12A), so, when A>B/9 so the radical is real and is also <1, there are 4 more solutions besides the two obvious t[sub]2[/sub]=4.5 and t[sub]5[/sub]=13.5
The solutions are:
t[sub]1[/sub]=(9/[&pi;])cos[sup]-1[/sup]([&radic;](0.75-B/12A))
t[sub]3[/sub]=(9/[&pi;])cos[sup]-1[/sup](-[&radic;](0.75-B/12A))
t[sub]4[/sub]=t[sub]1[/sub]+9
t[sub]6[/sub]=t[sub]3[/sub]+9
Unfortunately scully44, if you looked for simpler formulae, that's it. However it turns out that t[sub]1[/sub] tends to 1.5 when A/B goes towards infinite, that it is of course equal to 4.5 when A=B/9 and that it is already =~1.76 when A=B. Similarly t[sub]3[/sub] tends to 7.5 (starting again at 4.5) and is already =~7.24 when A=B.

prex

http://www.xcalcs.com

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[JamesBarlow]

I would have taken the easy way out.

I would have graphed it over 0<t<18 and see where it hits 0. This might have been tough if all you have available is a spreadsheet but most math programs could do this. Took a minute or two to perform this in Maple and then you can manipulate the numbers to better model the system instead of just looking a few numbers.

 

[rb1957]

prex made a little mistake in his factoring (at least the way i read his expression). inside the sqrt i started with 3A-B/3 = 3A/4-B/12 when you bring the "2" inside the sqrt; ok.

this = A*(3/4-B/12A), not (3/4-B/12A)

 

[prex]

Well rb1957, your expression
y*(-4y^2+(3A-B/3)) = 0
should be
y*(-4Ay^2+(3A-B/3)) = 0
so ...
It is instructive to follow this other way of reasoning: the cos() function results necessarily in a non dimensional number, so its argument must also be non dimensional. Now A and B may be dimensional (e.g. an amplitude, hence a length), therefore the quantity under the radical may only depend on the ratio A/B to be non dimensional...
It is BTW an evidence from the first equation presented by scully44 that the roots may only depend on A/B, not on their separate values.

prex

http://www.xcalcs.com

: Online tools for structural design

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[rb1957]

touche, another "small" mistake !

 

[pmover]

scully44,

hopefully the attached file will help you. In this file, A=B=1. So, you can change the values of A and B in cell B7.

You can use the Solver feature or maybe even the Goal Seek (doubt it) feature in Excel to further assist you.

Good Luck!
-pmover