Finding amplitude from superposition

[adi adi]

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.

 

[theodoros.mihos]

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

 

[adi adi]

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

pardon me i don't understand where does the cos come from?

 

[theodoros.mihos]

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

 

[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) .

 

[adi adi]

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 ?

 

[ehild]

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"

 

[adi adi]

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)

 

[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.

 

[adi adi]

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

sorry can you submit your work here? i got stuck

 

[adi adi]

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.

 

[theodoros.mihos]

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

 

[ehild]

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?

 

[theodoros.mihos]

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 $\phi$.

 

[BvU]

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.

 

[ehild]

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.

 

[adi adi]

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.

 

[ehild]

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?

 

[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 !

 

[ehild]

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.