# Undamped forced oscillator

• ChiralSuperfields
In summary, the conversation discusses a problem where the solution is to find whether an equation of the form 15.35 can satisfy equation 15.34 by plugging in suitable values for ##A, \omega, \phi##. It is found that 15.34 is satisfied by 15.35 under certain conditions for ##\phi## and ##A##, and any solution to the unforced system can also be added to produce another solution. The values of ##A## and ##\phi## were not initially defined, but were derived to match the given equation.
ChiralSuperfields
Homework Statement
Relevant Equations
For this problem,

The solution is,

However, can someone please explain how this is showing equation 15.35 as a solution of equation 15.34? I though both sides should be equal without assuming that ##\phi = 90##

Also why are they allowed to assume ##\phi = 90##?

Many thanks!

Given 15.34, we want to see whether an equation of the form 15.35 can be a solution if we plug in suitable values for ##A, \omega, \phi##.
To check this, we use 15.35 to substitute for x in 15.34.
It turns out that 15.34 is satisfied by 15.35 provided ##\phi=\pi/2+2n\pi## and ##A=\frac{F_0}{m\omega^2-k}##.

ChiralSuperfields
haruspex said:
Given 15.34, we want to see whether an equation of the form 15.35 can be a solution if we plug in suitable values for ##A, \omega, \phi##.
To check this, we use 15.35 to substitute for x in 15.34.
It turns out that 15.34 is satisfied by 15.35 provided ##\phi=\pi/2+2n\pi## and ##A=\frac{F_0}{m\omega^2-k}##.

and any solution to the unforced (or homogeneous or ##F_0=0##) system can be added to produce another solution as dictated by initial conditions.

ChiralSuperfields
hutchphd said:
and any solution to the unforced (or homogeneous or ##F_0=0##) system can be added to produce another solution as dictated by initial conditions.
Thank you @hutchphd , that is good to know!

hutchphd
Callumnc1 said:
I though both sides should be equal without assuming that ϕ=90

Also why are they allowed to assume ϕ=90?
Φ and A were not defined initially. They were testing a function to see what would be required for it it satisfy the original equation. Then they derived what Φ and A had to be for that form to be correct.

ChiralSuperfields
DaveE said:
Φ and A were not defined initially. They were testing a function to see what would be required for it it satisfy the original equation. Then they derived what Φ and A had to be for that form to be correct.
Oh ok thank you @DaveE that makes more sense now!

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