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

[randoreds]

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!

 

[Simon Bridge]

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.

 

[randoreds]

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!

 

[Simon Bridge]

Very often in science, it is not so much a question of getting the right answers as asking the right questions :)

 

[Nickie]

I would approach this problem by first understanding the concept of forced oscillations. Forced oscillations occur when an external force is applied to a system that is already oscillating. In this case, the external force is the sinusoidal force with an amplitude of 1.7 N.

To find the force extended, we need to use the equation for forced oscillations, which is ω^2= ω°^2 ± (( Fext/m)/A), where ω is the natural frequency of the system, ω° is the frequency of the driving force, Fext is the external force, m is the mass of the object, and A is the amplitude of the driving force.

Since we are given the amplitude (A = 1.7 N) and we need to find the force extended, we can rearrange the equation to solve for Fext. This gives us Fext = ω^2m(A-ω°^2). Plugging in the values given in the problem (m = 0.15 kg, A = 1.7 N, ω° = ?), we can solve for Fext.

Now, the second part of the problem asks us to find the frequency at which the force will make the object vibrate with an amplitude of 0.44 m. To do this, we can use the same equation, but this time we know the amplitude (A = 0.44 m) and we need to solve for ω°. Rearranging the equation, we get ω° = √((ω^2m(A-ω°^2))/Fext). Plugging in the values we found for Fext and the given values for m and A, we can solve for ω°.

I hope this helps to clarify how to find the force extended and the frequency in this problem. It is important to understand the concept of forced oscillations and how to use the equation to solve for different variables. Let me know if you have any further questions.