# Thread: Calculate the times where two waveforms cross

1. ## Calculate the times where two waveforms cross

Hi all,

The question I have is:

Calculate between 0 and 2.25s, the times at which the waveforms:
v=3cos3t and v=4sine6t cross.

So far I have calculated the time periods for both waveforms:

For Cosinewave: T = 2\pi /3 s

as v=3cos(3t) or v = 3cos(\omega t) where \omega = 2\pi f
Therefore: \omega = 2\pi /T = 3
So T = 2\pi /3

For Sinewave: T = 2\pi /6 s

as v=4sine(6t) or v = 4sine(\omega t) where \omega =2\pi f
Therefore: \omega = 2\pi /T = 6
So T = 2\pi /6

So it is evident that my sinewave is exactly twice the frequency of my Cosinewave, and
obviously what I want to achieve is to calculate the times where 3cos3t = 4sin6t.

I believe its required to use trigonometrical identoties to solve this problem from this point, but what approach/method to take I have no idea at this time.

Any help on this is greatly appriciated.

Thanks,
Dave

P.S: I am aware that I'm asking a question on my first post, but I intend to be of use to others here, where possible anyway!

2. Originally Posted by davet101
Hi all,

The question I have is:

Calculate between 0 and 2.25s, the times at which the waveforms:
v=3cos3t and v=4sine6t cross.

So far I have calculated the time periods for both waveforms:

For Cosinewave: T = 2\pi /3 s

as v=3cos(3t) or v = 3cos(\omega t) where \omega = 2\pi f
Therefore: \omega = 2\pi /T = 3
So T = 2\pi /3

For Sinewave: T = 2\pi /6 s

as v=4sine(6t) or v = 4sine(\omega t) where \omega =2\pi f
Therefore: \omega = 2\pi /T = 6
So T = 2\pi /6

So it is evident that my sinewave is exactly twice the frequency of my Cosinewave, and
obviously what I want to achieve is to calculate the times where 3cos3t = 4sin6t.

I believe its required to use trigonometrical identoties to solve this problem from this point, but what approach/method to take I have no idea at this time.

Any help on this is greatly appriciated.

Thanks,
Dave

P.S: I am aware that I'm asking a question on my first post, but I intend to be of use to others here, where possible anyway!
to solve $3\cos 3t=4\sin 6t$ use $\sin 6t=2\sin 3t \cos 3t$.
did this help?

3. Yes, thanks very much for the contribution.

However I still have some problems with this, sorry if this is me being a tad simple.

Using sine6t = 2sine3t cos3t I have achieved the following:

sine6t = 2sine3t cos3t

so 3cos3t = 4 * 2sine3t cos3t

so 3cos3t = 8sine3t cos3t

so 3 = 8sine3t

so 3/8 = sine3t

so (sine^-1(3/8))/3 = t = 0.128132 s (6.d.p)

As I have plotted the waveforms I am aware that they again intercest when the Cosinewave passes through the x axis, so:

when cos3t = 0

then t = (cos^1(0))/3 = pi/6 s

However, this is the first 2 occasions the waveforms intersect during the period between 0 and 2.25s, and I am required to calculate all the times at which the waveforms cross in this time which is 5 times (I have plotted the waveforms). How can the remaining value be obtained?

Thanks
Dave

4. Originally Posted by davet101
Yes, thanks very much for the contribution.

However I still have some problems with this, sorry if this is me being a tad simple.

Using sine6t = 2sine3t cos3t I have achieved the following:

sine6t = 2sine3t cos3t

so 3cos3t = 4 * 2sine3t cos3t

so 3cos3t = 8sine3t cos3t

so 3 = 8sine3t

so 3/8 = sine3t

so (sine^-1(3/8))/3 = t = 0.128132 s (6.d.p)

As I have plotted the waveforms I am aware that they again intercest when the Cosinewave passes through the x axis, so:

when cos3t = 0

then t = (cos^1(0))/3 = pi/6 s

However, this is the first 2 occasions the waveforms intersect during the period between 0 and 2.25s, and I am required to calculate all the times at which the waveforms cross in this time which is 5 times (I have plotted the waveforms). How can the remaining value be obtained?

Thanks
Dave
(shrugs) Then you missed one. Apart from verifying that (pi)/6 is a solution, I graphed it and found a solution at 0.15(pi) or so. Close enough for (pi)/6 for me.

-Dan

5. I dont think I explained my question very well there.

As you say I have obtained the first two points where the waveforms cross (at 0.128132s and pi/6s).

However this still leaves me neednig to find the remaining three solutions as I'm required to calulate all times they cross between 0 and 2.25s. I did think these could be evenly spaced but upon not very close inspection of my plot its obvious this is not true!

Any ideas how to obtain the remaining 3 solution? Sorry if this is already clear, I've been pulling my hair out over this question (believe it or not) and just have some mental block on it.

Dave.

6. Originally Posted by davet101
I dont think I explained my question very well there.

As you say I have obtained the first two points where the waveforms cross (at 0.128132s and pi/6s).

However this still leaves me neednig to find the remaining three solutions as I'm required to calulate all times they cross between 0 and 2.25s. I did think these could be evenly spaced but upon not very close inspection of my plot its obvious this is not true!

Any ideas how to obtain the remaining 3 solution? Sorry if this is already clear, I've been pulling my hair out over this question (believe it or not) and just have some mental block on it.

Dave.
Originally Posted by davet101
so 3/8 = sine3t
.
.
.
when cos3t = 0

then t = (cos^1(0))/3 = pi/6 s
First there are more solutions to cos(3t) = 0 than (pi)/6. $cos^{-1}(0) = \frac{\pi}{2}$ but bear in mind that the inverse cosine function is designed to only return one value. So what I am asking you is what are the solutions to cos(y) = 0? (Graph it if you can't think of any other way.) Then divide those answers by 3.

But that's not all! You have another equation you can work with: 3/8 = sin(3t). How can you solve that?

-Dan

7. Thanks for all your contribution guys, I have this sorted out now and my workings are below:

V1 = 3cos3t V2 = 4sine6t

Using trigonomic identity: sine(2A) = 2sineA cosA where A = 3t

So sine6t = 2sine3t cos3t (looking for when V1 = V2):

so 3cos3t = 4*2sine3t cos3t so 3cos3t = 8sine3t cos3t (equasion 1)

so 3 = 8sine3t so sine^-1(3/8) = 3t

so (sine^-1(3/8)/3)= t = 0.128132s (6.d.p)

Using: 3cos3t = 0:

so (cos^-1(0))/3 = t = pi6 s

so (pi-sine^-1(3/8))/3= t = 0.919065s (6.d.p)

so cos^-1(0) = 3t = pi/2 s

so (2pi + sine^-1(3/8))/3 = t = 2.222527 (6.d.p)

Thanks again

Dave