# Angular and Graphic Relationships for sine waves

Discussion in 'Homework Help' started by Ripplenwinder, Jan 17, 2015.

1. ### Ripplenwinder Thread Starter Member

Jan 17, 2015
5
0
?

Last edited: Jan 8, 2017
2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,558
517
V=Vm*sin(2*pi*f*t)
50=Vm*sin(2*pi*20,000*0.000005)
Find Vm.

3. ### MrChips Moderator

Oct 2, 2009
12,853
3,527
The question ignores the phase angle at t = 0.

v(t) = V sin(ωt + φ)

You can solve for V if you assume φ = 0.

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,558
517
V(t)=Vm*sin(2*pi*f*t)
You are given V(0.000005) (50), you are given f (20,000), you are given t (0.000005).
50 at time 0.000005=Vm*sin(2*pi*20,000*0.000005)

To find Vm all you have to do is:
$
Vm=\frac{V}{sin(2*pi*f*t)}
$

Just plug the eFing numbers!

5. ### WBahn Moderator

Mar 31, 2012
18,261
4,941
My guess is that this is to be inferred by calling it a "sine" wave (as opposed to a "cosine" wave, for instance). Authors often think that the two are somehow different since they are quite distinct in a trigonometry class but forget that when talking about sinusoidal waveforms that it is purely a distinction due to an arbitrary choice of time reference (and, arguably, even in trig it is a matter of arbitrary choice of reference axis, but that is much more highly standardized than time reference).

6. ### WBahn Moderator

Mar 31, 2012
18,261
4,941
It sounds like you might be well served just working on your algebra skills. That way you don't need to know a bunch of different formulas and variations on them, but rather can work with a few fundamental formulas and manipulate them to get what you need. In this case, you have

$
v(t) \; = \; V_m \sin$$2\pi f t$$
$

If we are wanting to find Vm, we just need to manipulate this equation to get Vm by itself on one side of the equation. We can do this by dividing both sides by sin(2πft) to get

$
V_m \; = \; \frac{v(t)}{\sin$$2\pi f t$$}
$

You are given that f = 20 kHz and that v(5μs) = 50V. This means that

$
V_m \; = \; \frac{50V}{\sin$$2\pi (20kHz)(5\mu s)$$}
$

the units on 2π in this case is radians/cycle and the units on Hz is cycles/second. After taking the scaling prefixes into account, we end up with

$
V_m \; = \; \frac{50V}{\sin$$0.2\pi$$} \; = \; \frac{50V}{0.5878} = 85.07V
$

Also, there is no need to report answers to 10 significant figures. As a general rule, engineering results are seldom better than 1%, or two sig figs, and so we routinely want our answers to be mathematically accurate to three sig figs. To help ensure this, intermediate results that are used in later calculations should be recorded to four or five sig figs to minimize accumulated round off errors. So the answer above would be reported as 85.1V.

7. ### WBahn Moderator

Mar 31, 2012
18,261
4,941
Just algebra. The basic concepts of algebra is very simple -- whatever you do to one side of an equation you do to the other. As long as you do the same thing to both sides (there are some caveats such as dividing by zero not being allowed), then the two sides remain equal. So, for example, if I want to find the value of x that makes the following equation true:

5x + 20 = 70

I can subtract 20 from both sides:

5x + 20 - 20 = 70 - 20
5x = 50

Then I can divide both sides by 5:

(5x/5) = (50/5)
(5/5)x = 10
x = 10

I could also do it the other way around and divide both sides by 5 first:

(5x + 20)/5 = (70/5)
(5x)/5 + (20/5) = 14
x + 4 = 14

And now I can subtract 4 from both sides:

x + 4 - 4 = 14 - 4
x = 10

In either case I am starting with 5x+20 and asking what simple steps can I take to progressively isolate the x.

I remember in sixth grade we were first learning formulas for things like area and perimeters of simple geometric shapes and we had for the area of a circle

$
A \; = \; \pi r^2
$

This was the first time I had ever encountered this. We also knew that the radius was the distance from the center of the circuit to the edge and that the diameter was the distance from edge to edge going through the center. It was pretty obvious that the diameter was twice the radius. At one point in class I went up to the instructor really excited because I had discovered a new formula, namely

$
A \; = \; \frac{\pi d^2}{4}
$

I had NOT just substituted r=d/2 into the formula, I had discovered it independently, probably by observing the behavior of some examples. I had no idea why the 4, specifically, was there. The instructor gave me an encouraging word and then showed me how my discovery was merely the result of substituting r=d/2 into the other formula. At the same time I was disappointed at having my discovery turn out to be not very new, but I had also just had my first glimpse into a whole new mathematical world, namely algebra.

You might pick up an Algebra I text -- you can find lots of them dirt cheap, even free for Kindle, on Amazon and the like -- and start working your way through it. You will be amazed at the doors it will open for you.

8. ### WBahn Moderator

Mar 31, 2012
18,261
4,941
Oh, no one here is ever obligated to reply or help -- we do so because we want to.

You've got quite a road ahead of you if you want to start playing with transform methods. You will need a good understanding of calculus and, preferably, differential equations (Laplace transforms are merely a very powerful and handy way for solving the differential equations that describe the behavior of linear circuits). The big key to calculus is a rock solid grounding in algebra. Trig is also very useful for a lot of calc and also a lot of circuit concepts. If you can work through analytic geometry before you tackle calculus I think you will find the effort worth it. Calc I and II are pretty major courses but Calc III is not as important to understanding differential equations. However, if you want to play with electromagnetics in any depth, then the material in Calc III comes into play.

Along the way you need to get very comfortable with exponentials and logarithms and also complex numbers and complex arithmetic.

You need to be backing all of this up with the fundamental physics involved in circuits, which is usually provided well enough by a good calculus-based Physics II course.

9. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,558
517
That is standard equation for sinusoidal signals. Your particular example has phase angle of zero. The full equation is the one MrChips posted:
V=Amplitude*sin(2*pi*f*t+phase angle)

Last edited: Jan 17, 2015
10. ### WBahn Moderator

Mar 31, 2012
18,261
4,941
Actually, MrChips posted that.

11. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,558
517
Thank you.

Sorry MrChips!

12. ### MrChips Moderator

Oct 2, 2009
12,853
3,527
@Ripplenwinder

If you are trying to get back into sine waves and trigonometry, think of a point rotating on the circumference of a circle in a counter-clockwise direction.

In the diagram above, the point P is radius a from the center of the circle.
As the point rotates in a counter-clockwise direction, the angle θ is increasing.

By trigonometry,

sin(θ) = y/a
or
y = a sin(θ)

Angular velocity ω describes how fast the angle θ is increasing. The units of ω are angle/time.
Angle is measured in radians, time in seconds.
Hence the units of angular velocity ω are radians/sec.

Now suppose the point P orbits around the center once every second, i.e. 360° in one second.
Hence the angular velocity ω = 2π rad/s.

In your question, the point rotates around the circle 20,000 times each second (20kHz).
Hence the angular velocity ω = 2πf = 2π x 20000 radians per second.

Another way of asking the question is how long would it take to go around the circle once?
The answer is 1/20 000 seconds = 1 000 000/20 000 μs = 50μs

So the next question is where would the point P be after 5μs if it began (t = 0) on the horizontal (θ = 0)
The answer is it would be 1/10 around the circle at θ = 36° or 2π/10 radians.

Now we can use
y = a sin(θ)

where θ = 36° or 2π/10 radians
y = 50
a = y/sin(θ) = 50/0.588 = 85

In summary, note that a sine wave is generated as the height y of the point P above the y=0 axis as P revolves around the center with an angular velocity ω.

Since

θ = ωt

and

ω = 2πf

we can write

y = a sin(θ)
y = a sin(ωt)
y = a sin(2πft)

Last edited: Jan 18, 2015
13. ### MrChips Moderator

Oct 2, 2009
12,853
3,527
Shucks. No, I am not using phasors, time or frequency domains, except the sine wave shown is in time domain. The circle is in spatial domain. i.e. imagine a wagon wheel rotating.

The constant between ω and f is 2π.

ω/f = 2π

Your questions are more complex than my answers. Maybe someone else would like to try and give you some answers.

14. ### MrChips Moderator

Oct 2, 2009
12,853
3,527
Ok. Forget about math and physics and algebra and geometry and trigonometry for a moment.

Take a real wagon wheel. No domains whatsoever.
Take an LED (and a battery to light the LED) and place in towards the rim of the wagon wheel.
Now rotate the wagon wheel so that the LED is now moving in a circle.
Now walk towards the front of the wagon so that you are looking at the wheel edgewise.
If you can still see the LED describe what you see.

Just remember to jack up the whole wagon so that it doesn't run over you.

Last edited: Jan 18, 2015
15. ### MrChips Moderator

Oct 2, 2009
12,853
3,527
Any kind of wheel will work. Anything that is circular and can rotate, bicycle, wheelchair, wheelbarrow, umbrella.

Do you know what is a thought experiment?