Trig function equation: d=4(sin5t-4cos6t)

[cpyykkonen]

Homework Statement

1. A weight hangs from a spring. If a force is applied to the weight at t = 0 seconds, it will start moving up and down. The following equation gives the distance d, in centimetres, of the weight from its equilibrium point: d=4(sin5t-4cos6t)

At what times during the first 2 seconds is the weight at the equilibrium point?

Homework Equations

The Attempt at a Solution

Since d=0 at equilibrium, we can set the equation 4(sin5t-4cos6t)=0
Since the factored out 4 will only affect the amplitude of the function, we can get rid of that (since we are looking for points of 0)
So I am left with sin5t-4cos6t=0
or sin5t=4cos6t
I understand that this is a difference of 2 functions (sint and cost), and the 0's will occur when each individual functions have equal but opposite magnitudes...I just have no idea how to proceed to find those points.

 

[Buzz Bloom]

Hi @cpyykkonen:

Here is an idea that you might find useful. Are you familiar with trig forms for the sum/difference of two angles? For example:
cos (x+y) = ?
sin (x-y) = ?

Regards,
Buzz

 

[cpyykkonen]

Yes, I am familiar with that. Duh, I never thought to try...I'll give it a go, thanks!

 

[haruspex]

Yes, I am familiar with that. Duh, I never thought to try...I'll give it a go, thanks!

I don't think that will get you very far. Looks like it will produce a twelfth order polynomial, or some such.
I wouid start with sketching it, through a spreadsheet, say.

 

[SammyS]

I don't think that will get you very far. Looks like it will produce a twelfth order polynomial, or some such.
I wouid start with sketching it, through a spreadsheet, say.

... or use a graphing calculator.

 

[cpyykkonen]

Yeah it got out of control pretty quick, using the angle sum/difference identities. I can graph it through Desmos and find the zeros that way...but it doesn't help with my assignment. Teacher is looking an algebraic solution, with an equation to represent the zeros.

 

[haruspex]

Yeah it got out of control pretty quick, using the angle sum/difference identities. I can graph it through Desmos and find the zeros that way...but it doesn't help with my assignment. Teacher is looking an algebraic solution, with an equation to represent the zeros.

Not that it helps much, but are you sure it isn't 4sin(5t)- 4cos(6t)?

 

[cpyykkonen]

Yes, I just double checked. The original equation is d=4(sin5t-4cos6t)

 

[Buzz Bloom]

I don't think that will get you very far. Looks like it will produce a twelfth order polynomial, or some such.

Teacher is looking an algebraic solution, with an equation to represent the zeros.

Hi cpyykkonen and haruspex:

I have not done the detail work, but I think you might be able to get a sixth order equation in sin t. But even if it is a twelfth order equation, why doesn't that satisfy the Teacher's requirement that it represent the zeros? I understand a sixth or twelfth is not likely to have a closed solution, but perhaps that is not what the Teacher is looking for.

Here are some thoughts about the degree of the equation.
6t = t + 5t
sin 5t = fifth order polynomial in sin t, all terms odd powers.
cos 5t = fifth order polynomial in cos t, all terms odd powers.

Regards,
Buzz

 

[Ray Vickson]

Yes, I just double checked. The original equation is d=4(sin5t-4cos6t)

Are you really sure you want an algebraic solution? If so, here is the one I got using Maple. (Normally I would not supply a solution, as this goes against PF policy, but I doubt the exact solution would be of any use to you, especially as it is just stated without proof.)

To find the smallest positive root $t_1$ of your equation, proceed as follows. Let $z_0$ be the 6th root of the 12th degree polynomial
$$p(z) = 16384 z^{12}-48896 z^{10}+54656 z^{8}-28112 z^{6}+6520 z^{4}-551 z^2+15$$ (where the indexing of the roots of a polynomial are given by a particular rule in Maple, a rule I will not state here).
Then
$$t_1 = \arctan\left(\frac{2048 z_0^{10} -5600 z_0^{8} +5560 z_0^{6} -2410 z_0^{4} +425 z_0^2 - 19}{\sqrt{z_0}} \right) \doteq 0.223974 $$ I guess your teacher wants you to deduce and prove a formula like the above in order to avoid using numerical methods of solution. However, just about any expert you consult would suggest the opposite; that is, they would suggest numerical method of solution and would suggest that you forget even trying to solve it "symbolically".

Similar formulas can be given for the other three positive roots that are less than 2.

 

[SammyS]

Homework Statement

1. A weight hangs from a spring. If a force is applied to the weight at t = 0 seconds, it will start moving up and down. The following equation gives the distance d, in centimetres, of the weight from its equilibrium point: d=4(sin5t-4cos6t)

At what times during the first 2 seconds is the weight at the equilibrium point?
...

The Attempt at a Solution

Since d=0 at equilibrium, we can set the equation 4(sin5t-4cos6t)=0
Since the factored out 4 will only affect the amplitude of the function, we can get rid of that (since we are looking for points of 0)
So I am left with sin5t-4cos6t=0
....

Not that it helps much, but are you sure it isn't 4sin(5t)- 4cos(6t)?

The following may be of some help, although I doubt it gives a closed form solution.

$\sin(5 t) - \cos(6 t) = 1 - 2 \cos^2(3 t) + \sin(5 t)$

$\sin(5 t) - \cos(6 t) = -1 + 2 \sin^2(3 t) +\sin(5 t)$

Use either to get $\ 4\left( \sin(5 t) - \cos(6 t) \right) \$ and then subtract $\ 3\sin(5t) \ .$

 

[cpyykkonen]

Thanks for all of your replies. This question is getting a little over my head to solve it algebraically. I graphed each functions (sin5t) and (4cos6t) and found the intersecting point, which works out to the 0's. This worked out to 0.224, 0.76, 1.3, 1.842 seconds (during the first 2 second period). I think I'll submit my work this way...and just keep working on the problem in my own time! Gotta get the assignment in!

 

[haruspex]

Thanks for all of your replies. This question is getting a little over my head to solve it algebraically. I graphed each functions (sin5t) and (4cos6t) and found the intersecting point, which works out to the 0's. This worked out to 0.224, 0.76, 1.3, 1.842 seconds (during the first 2 second period). I think I'll submit my work this way...and just keep working on the problem in my own time! Gotta get the assignment in!

Those are the numbers I got too.