Undamped forced oscillator

  • Thread starter ChiralSuperfields
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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.
  • #1
ChiralSuperfields
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Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1675720881858.png

The solution is,
1675720909212.png

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!
 
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  • #2
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}##.
 
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  • #3
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}##.
Thank you for your reply @haruspex !
 
  • #4
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.
 
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  • #5
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!
 
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  • #6
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.
 
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  • #7
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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