# Finding amplitude from superposition

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1. May 29, 2015

1. The problem statement, all variables and given/known data
so there is this problem : find the amplitude from the superposition of Z1 and Z2 where
Z1 : 8 sin 100t, A1= 8m
and Z2 : 6 sin (100t-pi/2), A2=6m

2. Relevant equations
i know that all we need to do is add them and do the trig using sine additon, but i couldnt do that because the amplitude is different.
3. The attempt at a solution
yeah they give me the key answer but i dont understand because the solution only stated that all you need to do was do pythagorean theorem based on each amplitude.

2. May 29, 2015

### theodoros.mihos

$$\sin(100t-\pi/2) + \cos(100t) = ???$$

3. May 29, 2015

pardon me i dont understand where does the cos come from?

4. May 29, 2015

### theodoros.mihos

Argumets are differs by $\pi/2$, and functions are orthogonal, thus... Pythagoras.

5. May 29, 2015

### ehild

What do you get if you apply the addition law to sin(100t-pi/2)?

Or even simpler: You know that sin(pi/2-x) = cos(x) And sin(-x)=sin(x)

Edit: I think I was sleeping... Of course, sin(-x)=-sin(x) .

Last edited: May 29, 2015
6. May 29, 2015

sorry isnt sin -(x) = - sin x ?

7. May 29, 2015

### ehild

What do you mean with sin - (x)? Sine must have an argument. It is a function. You can not subtract something from "sin"

8. May 29, 2015

sorry i mean sin(-x) = -sin(x)

9. May 29, 2015

### ehild

So sin(-x)=-sin(x). sin(x) is an even function. sin(pi/2)=1 and sin(-pi/2) = -1.
Edit: Still sleeping, and forgetting English.
Of course sine is odd.

Last edited: May 29, 2015
10. May 29, 2015

sorry can you submit your work here? i got stuck

11. May 29, 2015

yeah thank you, so how do you solve this case?
i change the second wave function (Z2) from 6sin(100t-pi/2) to -6cos(100t), so i have this problem :
8sin(100t) - 6cos(100t), and i got stuck right there.

12. May 29, 2015

### theodoros.mihos

Energy conservation: $$A_{12}^2 = A_1^2 + A_2^2$$ because there are orthogonal.

Last edited: May 29, 2015
13. May 29, 2015

### ehild

You have to write 8 sin(100t)-6cos(100t)in the form A sin(100t + Φ ) where Φ is the phase constant. You can expand it, applying the addition rule for sine. What do you get?
The equation has to be valid at every time, when 100t = 0 and when 100t=pi/2. What equations you get for Φ?

14. May 29, 2015

### theodoros.mihos

This is the full solution that gives you two equation for $A$ and $\phi$.

15. May 29, 2015

### BvU

Ehild ! sin(x) is an odd function !
And Adi2 only asked in post 6 because of the typo in your post 5: the sin(-x)=sin(x) of course was meant to be $\sin(-x)= -\sin x$

Adi: I always like to make a picture. Better than many words and useful to check the outcome.

16. May 29, 2015

### ehild

Thank you BvU to pointing out my errors. I must have been very confused this morning...

I wanted to make the OP recognize that A sin(Φ) = -6 and Asin(pi/2+Φ) = Acos(Φ) = 8, and he can get A2 by adding the squares.

Last edited: May 29, 2015
17. May 31, 2015

so i get this : A (sin100tcosΦ + sinΦcos100t)
when 100t=0, i get this : A sin Φ
when 100t=pi/w, i get this : A cos Φ
so how do i find the Φ ? thank you for being patient because im veeery slow at trigs.

18. May 31, 2015

### ehild

Remember it was 8 sin(100t)-6cos(100t)=A sin(100t + Φ )
What is the right side when 100t=0? What is it when 100t=pi/2? So what equations hold for cosΦ and sinΦ?

Last edited: May 31, 2015
19. May 31, 2015

### BvU

Oh boy, here's me interfering again. So sorry!

Adi has already shown what the right-hand side is, so now he/she should evaluate the left-hand side .

But both Ehild and I mean the same thing: "I get $A\sin\phi$" is incomplete: you were intended to get $... = A\sin\phi$ with the emphasis on the = sign.
The $A\sin\phi$ is a value, but $... = A\sin\phi$ is an equation !

20. May 31, 2015

### ehild

Yes, I meant equations when 100t = 0 and when 100t = pi/2. What equations you get for Φ? (I asked in Posts #13 and #18). You got the right sides, but they must be equal to the left sides.