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Sinusoid question

  • Thread starter ecas
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  • #1
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Homework Statement


Find the amplitude and phase of the sinusoid (4√3-3)cos(2t + 30°) + (3√3 – 4)cos(2t + 60°)


Homework Equations


Acos wt + Bsin wt


The Attempt at a Solution


Answer should be amplitude 5 and phase 36.9° but I get 4.11 and 5.7
 

Answers and Replies

  • #2
berkeman
Mentor
56,648
6,549

Homework Statement


Find the amplitude and phase of the sinusoid (4√3-3)cos(2t + 30°) + (3√3 – 4)cos(2t + 60°)


Homework Equations


Acos wt + Bsin wt


The Attempt at a Solution


Answer should be amplitude 5 and phase 36.9° but I get 4.11 and 5.7
Welcome to the PF. Please show us your work, so that we can try to help you. That's how it works here on the PF.
 
  • #3
5
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The square root of (4√3-3)^2 + (3√3 – 4)^ 2 = 4.11 and θ = tan^-1 (4√3-3)/(3√3 – 4) = 16.9 degrees
 
  • #4
berkeman
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The square root of (4√3-3)^2 + (3√3 – 4)^ 2 = 4.11 and θ = tan^-1 (4√3-3)/(3√3 – 4) = 16.9 degrees
Not sure what that's meant to represent. You have two sinusoidal functions of time, which can be added to make a new sinusiodal function, with a combined amplitude, and a combined resultant phase. It should look something like this:

f(t) = A cos(2t + phase)

What would be a way that you can add two cosine functions that have different phases?
 
  • #5
berkeman
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  • #6
5
0
I tried changing the second function from cos to sin to match the identity but this does not seem to affect the magnitude in any way. I still do not know how to arrive at A=5.
 
  • #7
The Electrician
Gold Member
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Does this have anything to do with the study of electronics? If so, have you studied phasor arithmetic yet?
 
  • #8
5
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Actually the problem comes before the section on phasors. Supposedly it is supposed to be solved with the trig identities we are given.
 
  • #9
Redbelly98
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See the link berkeman gave in post #5. You eventually want to get things in the form

A cos(2t) + B sin(2t)​

To do that you'll need to use the addition identity for cosine,

cos(x+y) = ???​
 
  • #10
5
0
Ok I am getting 5.1 which sounds right. Thanks everyone.
 

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