Frequency of Oscillation, with mass and k value given

AI Thread Summary
A mass of 2.0 kg attached to a spring with a force constant of 990 N/m oscillates with a frequency of approximately 3.5 Hz, calculated using the formulas for angular frequency and frequency. The discrepancy with the expected answer of 4.0 Hz was clarified as a possible rounding issue or an error in the answer key. The calculations were confirmed to be correct, and the approximation was acknowledged by the instructor. The discussion highlights the importance of understanding significant figures in physics problems. Overall, the participant felt reassured about their understanding of the concepts before their final exam.
LaLaLina
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Homework Statement



A mass m = 2.0 kg is attached to a spring having a force constant k = 990 N/m as in the figure. The mass is displaced from its equilibrium position and released. Its frequency of oscillation (in Hz) is approximately _____ . [/B]
upload_2014-12-5_17-52-8.png


Homework Equations


\omega=\sqrt\frac{k}{m}
T=\frac{2\pi}{\omega} f=\frac{\omega}{2\pi}
[/B]

The Attempt at a Solution



m= 2.0 kg
k= 990 N/m
f = ?
omega = sqrt((990N/m)/(2.0kg)) = 22.25 rad/s
f= (22.25 rad/s)/(2pi) = 3.5 Hz

The answer to the problem, is 4.0 Hz. However, I don't believe I should have rounded up. I believe I have made an error in how to solve the problem.

Have I gone wrong in choosing the formulas to use? or
Should I have taken something else into account based on the mention of the equilibrium statement? and if so, what does that imply?
I recall doing another type of spring problem where it went past its point of equilibrium, but we were using velocity and its angle and solving for v using
dy/dt = -A(omega)sin(omega(t)+phi), however I was reviewing the slides before this problem, and it doesn't seem to address the equilibrium.

I thank you for your assistance very much. I'm studying for my final on Monday and I'm trying to be prepared.[/B]
 
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Hi LaLaLina, Welcome to Physics Forums.

Your solution method and result look fine. It's possible that at some point someone changed a parameter in the problem in order to make it a "new" problem, but didn't update the answer key.
 
your calculation looks fine to me. 3.5 Hz is correct to 2 significant figures and 4 Hz is correct to 1 significant figure. Although, it is weird that your teacher (or professor or book) wrote 4.0 Hz, since this would usually imply 2 significant figures of precision... maybe it is just an error in the book like gneill says
 
Thank you so much. I got an email back from him, and he said I did wonderful. That it was supposed to be an approximation. So, there in lies the answer I guess. I on the other hand am sitting here worrying my brain out over knowing how to do a problem correctly with this final coming up that is 30% of my grade, so I wanted to make SURE I knew what I was doing.

Thank you both for your feedback.I feel much better now.
 
Glad we could help. That's what Physics Forums is all about :)
 
  • Like
Likes BruceW and LaLaLina
Oh my! I just now noticed the T.A.R.D.I.S.
fellow Whovian in the house!

Thanks again.
 
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