Finding amplitude from superposition

AI Thread Summary
The discussion revolves around finding the amplitude from the superposition of two wave functions, Z1 and Z2. Participants clarify that Z2 can be rewritten using trigonometric identities, specifically converting it from sine to cosine. The key to solving the problem lies in expressing the combined wave function in the form A sin(100t + Φ), allowing for the application of the Pythagorean theorem due to the orthogonality of the sine and cosine functions. The conversation emphasizes the importance of correctly identifying the phase constant and using equations derived from evaluating the wave function at specific time intervals. Ultimately, the focus is on understanding the mathematical relationships between the components of the wave functions to determine the resultant amplitude.
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Homework Statement


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

Homework Equations


i know that all we need to do is add them and do the trig using sine additon, but i couldn't do that because the amplitude is different.

The Attempt at a Solution


yeah they give me the key answer but i don't understand because the solution only stated that all you need to do was do pythagorean theorem based on each amplitude.
 
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$$ \sin(100t-\pi/2) + \cos(100t) = ? $$
 
theodoros.mihos said:
$$ \sin(100t-\pi/2) + \cos(100t) = ? $$
pardon me i don't understand where does the cos come from?
 
Argumets are differs by ##\pi/2##, and functions are orthogonal, thus... Pythagoras.
 
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) .
 
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ehild said:
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)
sorry isn't sin -(x) = - sin x ?
 
adi adi said:
sorry isn't sin -(x) = - sin x ?

What do you mean with sin - (x)? Sine must have an argument. It is a function. You can not subtract something from "sin"
 
ehild said:
What do you mean with sin - (x)? Sine must have an argument. It is a function. You can not subtract something from "sin"
sorry i mean sin(-x) = -sin(x)
 
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.
 
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  • #10
theodoros.mihos said:
Argumets are differs by ##\pi/2##, and functions are orthogonal, thus... Pythagoras.
sorry can you submit your work here? i got stuck
 
  • #11
ehild said:
So sin(-x)=-sin(x). sin(x) is an even function. sin(pi/2)=1 and sin(-pi/2) = -1.
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
Energy conservation: $$ A_{12}^2 = A_1^2 + A_2^2 $$ because there are orthogonal.
 
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  • #13
adi adi said:
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.
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
ehild said:
You have to write 8 sin(100t)-6cos(100t)in the form A sin(100t + Φ ) where Φ is the phase constant.
This is the full solution that gives you two equation for ##A## and ##\phi##.
 
  • #15
ehild said:
So sin(-x)=-sin(x). sin(x) is an even function. sin(pi/2)=1 and sin(-pi/2) = -1.
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.

Sines_addition.jpg
 
  • #16
BvU said:
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##

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.
 
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  • #17
ehild said:
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?
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 I am veeery slow at trigs.
 
  • #18
adi adi said:
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 I am veeery slow at trigs.
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?
 
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  • #19
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 :smile:.

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
adi adi said:
when 100t=0, i get this : A sin Φ
when 100t=pi/w, i get this : A cos Φ

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.
 
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