- #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

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- Thread starter ecas
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- #1

- 5

- 0

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

Acos wt + Bsin wt

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

- #2

berkeman

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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

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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http://www.cs.sfu.ca/~tamaras/sinusoids318/Adding_two_sinusoids.html

Are you sure you have the original equations in the right format to apply the equations?

- #6

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- #7

The Electrician

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- #8

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- #9

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

cos(*x+y*) = ???

- #10

- 5

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Ok I am getting 5.1 which sounds right. Thanks everyone.

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