Superposition of harmonic oscillations

[Lengalicious]

Homework Statement

Find the amplitude and phase shift of the following two superposed harmonic oscillations.

Homework Equations

x1(t)=3sin(2∏t+∏/4)
x2(t)=3cos(2∏t)

The Attempt at a Solution

Ok normally i would be able to do this, however one oscillation is cos and the other sin, so i can't use the trig identity for sin+sin or cos+cos. Is it possible to turn 3sin(2∏t+∏/4) into 3cos(2∏t+∏/4+∏/2) or is that invalid?

 

[Modulated]

Homework Statement

Find the amplitude and phase shift of the following two superposed harmonic oscillations.

Homework Equations

x1(t)=3sin(2∏t+∏/4)
x2(t)=3cos(2∏t)

The Attempt at a Solution

Ok normally i would be able to do this, however one oscillation is cos and the other sin, so i can't use the trig identity for sin+sin or cos+cos. Is it possible to turn 3sin(2∏t+∏/4) into 3cos(2∏t+∏/4+∏/2) or is that invalid?

I think you mean -π/2 there, but otherwise yep, that's fine. (Or there are similar trig identities you could use, but of course it will add up to the same thing.)

 

[Lengalicious]

Yeh -∏/2 is what i meant, one more question when calculating the amplitude i find that somehow the following equation is derived, A=√(A₁²+A₂²+2A₁A₂cos(θ₂-θ₁)). I don't understand how this is found?

 

[Lengalicious]

Bump: How do I find the amplitude, i get a phase of pi/8 and an amplitude of 5.5 but i think my amplitude is wrong, can someone check please?

 

[gneill]

Yeh -∏/2 is what i meant, one more question when calculating the amplitude i find that somehow the following equation is derived, A=√(A₁²+A₂²+2A₁A₂cos(θ₂-θ₁)). I don't understand how this is found?

Bump: How do I find the amplitude, i get a phase of pi/8 and an amplitude of 5.5 but i think my amplitude is wrong, can someone check please?

The formula above is an application of the cosine law. You can see how it applies if you consider that since both have the same frequency $(2\pi)$, the two functions can be represented in phasor form and added as phasors.

Convert the first sin function to cosine as you have suggested previously. That makes its phase angle $-\pi/4$. Its phasor is then a vector of length 3 with angle $-\pi/4$ from the x-axis. The second function has no phase, so it's a vector of length 3 on the x-axis. Add as you would any two vectors. You'll find that one way to find the magnitude is to apply the cosine law -- draw it, you'll see. You could also add them via the component method. Either works just fine.

 

[Lengalicious]

Ah, that helps a lot thanks very much