How do you find the force extended if given the amplitude? Forced Oscillations

AI Thread Summary
To find the force extended given the amplitude in a forced oscillation scenario, the equation ω² = ω°² ± (Fext/m)/A is used, where ω is the natural frequency, ω° is the driving force frequency, and A is the amplitude. In this case, the amplitude of 0.44 meters should be plugged into the equation to solve for the external force. The discussion highlights the importance of understanding the role of external forces in the motion of oscillating systems. Additionally, the distinction between transient and steady-state behavior is noted, with emphasis on using appropriate equations for each situation. The conversation concludes with a realization that asking the right questions is crucial in scientific problem-solving.
randoreds
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How do you find the force extended if given the amplitude? Is main questions, I also have one slight question.

Ok, doing a problem. There is no damping. A (.15kg) object is hanging from a light(6.30N/m) spring.

A sinusoidal force with an amp of 1.7 N drives the system. And the problem is asking at what frequency will the force make the object vibrate with an amp of .44m

So I'm solving and my final equation is ω^2= w°^2 plus or minus (( Fext/m)/A)
Where ω = natural frequency and ω°= frequency of the driving force.

So my questions are how would I find the force extended given the amplitude and ω^2= w°^2 plus or minus (( Fext/m)/A(------ Would I plug in .44 for that A?

Thanks in advance!
 
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What does "force extended" mean?
What role does it play in the motion?

Is the situation one that is transient or steady-state?
Do you have equations for those situations in your notes?

note: ##F_{ext}## is usually notation for an external force... just saying.
 
Last edited:
Simon Bridge said:
What does "force extended" mean?
What role does it play in the motion?

Is the situation one that is transient or steady-state?
Do you have equations for those situations in your notes?

note: ##F_{ext}## is usually notation for an external force... just saying.

Thanks, I can't believe I didn't think of that! I got the right answer.That took way longer than it should have!
 
Very often in science, it is not so much a question of getting the right answers as asking the right questions :)
 
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