Simple question regarding higher frequency vs lower frequency

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Homework Help Overview

The problem involves a 0.145 kg object attached to a spring with a spring constant of 6.30 N/m, subjected to a sinusoidal driving force. The objective is to determine the frequency at which the object will vibrate with a specified amplitude, considering both higher and lower frequency solutions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of finding the frequency, noting that the amplitude equation involves squared frequency terms. There is a focus on the need to consider both positive and negative square roots when solving for frequency.

Discussion Status

The discussion has progressed with the original poster successfully identifying the higher frequency solution. There is acknowledgment of a mistake regarding the consideration of the negative square root, which has been clarified through participant interaction.

Contextual Notes

The original poster is required to find both higher and lower frequency solutions, which introduces complexity in the problem-solving process. There is mention of a potential issue with email notifications for replies in the forum.

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Homework Statement



Damping is negligible for a 0.145 kg object hanging from a light 6.30 N/m spring. A sinusoidal force with an amplitude of 1.70 N drives the system. At what frequency will the force make the object vibrate with an amplitude of 0.440 m?

Homework Equations



I don't know how to put formulas in here but basically find out the initial omega value, then plug it into the forced oscillations amplitude equation to get the angular frequency. Then use the frequency formula to get your frequency.

The Attempt at a Solution



Here's the problem!

The question asks me for 2 solutions -- higher frequency and lower frequency.

I obtained the higher frequency given the equations above and it is indeed the correct answer. That answer is 1.33 Hz.

Now how do I get the lower frequency? It's not -1.33 if anyone is wondering.

Thanks for the help guys!
 
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Welcome to PF, Doc Z.

It's been a while since I solved this type of problem. But I remember that the amplitude equation involves frequency terms that are squared.

When you worked through your solution, you probably took a square root at some point. Did you consider the negative square root as well?
 
Sorry for the late response, for some reason I didn't get an email when I receive a reply to a thread.

Thank you very much! It was just a stupid mistake on my part, I completely forgot about the negative sign.
 
Doc Z said:
... for some reason I didn't get an email when I receive a reply to a thread.

Yeah, that happens sometimes. It seems to be a bug in the system.

If you click on "My PF" at the top left of the page, you'll see a list of recent replies in threads you are subscribed to. I have abandoned email notification in favor of this other way of checking for recent replies.

Glad the HW problem worked out.
 

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