Trouble figuring out what to do next, please help.

$Avatar for efsfsf\\eeeeeee$

If an external force F sin ωt is imposed on a system, the response is given by A sin(ωt + ϕ),
with ϕ the phase shift.
How would we express the response A sin(ωt + ϕ) in terms of sin ωt and cos ωt?

Using compound angle identities,
A sin(ωt + ϕ) = A(sin(ωt) cos ϕ + cos(ωt) sin ϕ)
= A cos ϕ sin(ωt) + A sin ϕ cos(ωt)

Stuck after this step.

Reply 1

mqb2766

Original post by efsfsf\\eeeeeee

If an external force F sin ωt is imposed on a system, the response is given by A sin(ωt + ϕ),
with ϕ the phase shift.
How would we express the response A sin(ωt + ϕ) in terms of sin ωt and cos ωt?
Using compound angle identities,
A sin(ωt + ϕ) = A(sin(ωt) cos ϕ + cos(ωt) sin ϕ)
= A cos ϕ sin(ωt) + A sin ϕ cos(ωt)
Stuck after this step.

Youve pretty much done it.
Acos(ϕ)
is the coefficient of sin(ωt) and similarly for the other term. The coefficients do not depend on t.

(edited 1 week ago)

Reply 2

$Avatar for efsfsf\\eeeeeee$

efsfsf\\eeeeeee

Original post by mqb2766

Youve pretty much done it.
Acos(ϕ)
is the coefficient of sin(ωt) and similarly for the other term. The coefficients do not depend on t.

Can you elaborate on how Acos(phi) is the coefficient?

Also the solution continues to do this, which is why I'm confused:

https://ibb.co/xmwQZNm

Reply 3

mqb2766

Original post by efsfsf\\eeeeeee

Can you elaborate on how Acos(phi) is the coefficient?
Also the solution continues to do this, which is why I'm confused:
https://ibb.co/xmwQZNm

As phi is a constant, then Acos(phi) is the constant which mutiplies the sinusoid sin(wt).

Its really just showing that the two representations are equivalent so
A sin(ωt + ϕ) <-> Bsin(ωt) + Ccos(ωt)
where B = Acos(ϕ) and C=Asin(ϕ). The previous work goes from left to right.

Now going from right to left, using pythagoras identity to remove the dependence on ϕ
B^2 + C^2 = A^2 cos((ϕ)^2 + A^2 sin(ϕ)^2 = A^2 (cos((ϕ)^2 + sin(ϕ)^2) = A^2
so
A = sqrt(B^2 + C^2)
Also dividing to remove the dependence on A
C/B = Asin(ϕ) / Acos(ϕ) = tan(ϕ)
so
ϕ = arctan(C/B)
For the usual range associated with arctan(), otherwise a bit more reasoning needs to be done.