good ,good, good,good,vibrations (not)

[scotsfool]

I f I can have these solved I would be a happy man.......

two vibrations, x1 = 3sin (10t + pi/6) and x2 = 2 cos (10t - pi/6) where t is in seconds, are superimposed. Determine the time at which the amplitude of the resultant vibration, x1 + x2, first reaches a value of 2.

can anyone solve?? I must be a fool cause I been over it many times and maybe that is the problem!!!!!

please help.

[Rassis]

t = 4.966

[scotsfool]

Thanks for the reply but I really want to see the way you got to the result, this would help greatly.

also can you write in english if possible :0)

I thank you for your patience and help

can anyone help? and translate

[krtxmrtz]

Thanks for the reply but I really want to see the way you got to the result, this would help greatly.

also can you write in english if possible :0)

I thank you for your patience and help

can anyone help? and translate

Probably he used numerical methods. As far as I know you can't invert x₁ + x₂
One easy way to find a solution is by means of a bisection algorithm.

[scotsfool]

Thanks, but I would like to see how it is done as I am at a loss with it....

[krtxmrtz]

Thanks, but I would like to see how it is done as I am at a loss with it....

Well, if you don't have a programmable handheld calculator then you may try the bisection method.

Call g(t) = x₁(t) + x₂(t)

Finding a value of t such that

g(t) = 2

is equivalent to finding a value of t such that

f(t) = g(t) - 2 = x₁(t) + x₂(t) - 2 = 0

Essentially you find "by eye" i.e. by trial and error 2 values of t, call them t₁ and t₂, such that f₁ * f₂ < 0

where I have called

f₁ = f(t₁)
f₂ = f(t₂)

i.e. either f₁ is negative and f₂ positive or the other way round. In any case this means that there must be at least one solution inside the interval (t₁, t₂) because f(t) is a continuous function.

Assume that t₂ > t₁. What you do now is calculate the middle point t_m of the interval:

t_m = (t₁ + t₂) / 2

and calculate the function at this point, f_m = f(t_m). If f_m has the same sign as f₁ then the solution must be in the interval (t_m, t₂) and you now assign t₁ = t_m and go back and loop. Or, if f_m has the same sign as f₂ then the solution must be in the interval (t₁, t_m) and now you let t₂ = t_m.

So every time you repeat (iterate) you end up with a smaller interval and a f_m nearer to 0. Eventually your f_m will be as small as you wish -as the accuracy of your calculator / spreadsheet or whatever allows you to calculate. At that point you can stop and take t_m as your solution.

[krtxmrtz]

Also you can try a "brute force" method. Maybe it's not so elegant but the aim is to find the solution.
Once you have found the 2 values of t bracketing the solution, t1 and t2, you can simply "buy as many tickets as necessary to win the lottery", in other words:

Sub Solve()
   Randomize
   'Set a conveniently small value
   fThreshold = 0.0000001
   'fun is your x1 + x2 - 2
   fMin = Min(Abs(fun(t1)), Abs(fun(t2)))
   Do
      t = t1 + Rnd*(t2 - t1)
      f = Abs(fun(t))
      If f < fMin Then fMin = f
   Loop While fMin > fThreshold
End Sub
Function fun(p)
   '(Use doubles if needed)
   Dim Pi6 As Single
   Dim p10 As Single
   Const Pi = 3.141593
   Pi6 = Pi/6
   p10 = 10*p
   fun = 3*sin(p10 + Pi6) + 2*cos(p10 - Pi6) - 2
End Function

I haven't tested this code but I think it should work.

[scotsfool]

that is above my level of intelligence!!!!!!!! very confused now

[Rassis]

I subscribe everything Krtxmrtz said. You can find the solution by trial and error using a spread sheet but it might take you quite a while, depending on the precision needed, or you can use the GOAL-SEEK option in EXCEL instead. In this case it will be quite straight forward. Please see the attached EXCEL file where I followed the following steps:

1. Construct a table with x1+x2 as a function of t departing from t = 0;
2. Construct a line graph;
3. Notice that, when t = 0, x1+x2 equals 3.23 which is > 2;
4. Notice also that as t increases x1+x2 also increases and starts decreasing when t is approximately to 0.1;
5. Notice finally that when t = 0.24, x1+x2 equals approximately 0, meaning that the first time x1+x2 reaches a value of 2 is somewhere between t = 0 and t = 0.24;
5. Enter in cell F2 any number comprised in the before mentioned interval;
6. Use GOAL-SEEK the way showed in the two images in sheet “Images” and that is all.

I am sorry for having given the answer 4.966. It is correct though because it returns x1+x2 = 2, but it is not the moment when x1+x2 reaches the value of 2 for the first time. I was not aware of the shape of the specific function at the time and just dropped any number in cell F2 (which I cannot remember anymore).

Rui

[krtxmrtz]

that is above my level of intelligence!!!!!!!! very confused now

Sorry if my explanation was confusing. I should have attached a simple drawing in the first place. Well here you go.

You can see that the points t₁ and t₂ bracket the solution because the function f is negative at t₁ and positive at t₂. Therefore, you find the midpoint t_m in between them and evaluate the function there. In the sample graph it happens to be positive so now you know the solution is between t₁ and t_m. Therefore you forget about t₂ and now you set t_m as your new t₂. This is your first step. Now you start all over again (of course, you must have the computer do it for you) and repeat over and over. The (t₁, t₂) interval becomes ever smaller and your solution in sandwiched in it.