Angular frequency of oscillation

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Homework Help Overview

The discussion revolves around the angular frequency of oscillation for a mass-spring system, specifically focusing on the relationship \(\omega \geq \sqrt{\frac{k}{m}}\), where \(\omega\) is the angular frequency, \(k\) is the spring constant, and \(m\) is the mass. The original poster seeks clarification on the meaning of the inequality in the context of simple harmonic motion.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the relationship using a differential equation approach and questions the significance of the \(\geq\) sign. Some participants suggest that under normal conditions, the relationship should be an equality rather than an inequality.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the inequality and the conditions under which it might apply. Some have provided insights regarding the linearity of the spring and the role of gravity in the system, indicating a productive examination of the assumptions involved.

Contextual Notes

Participants note that the problem specifies simple harmonic motion, which typically implies a linear spring behavior. There is also mention of a potential scenario where the spring's behavior could change, affecting the frequency, but this is not aligned with the conditions of the problem as stated.

jdstokes
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A mass m is suspended vertically by a spring of force constant k. Derive the relation [itex]\omega \geq \sqrt{\frac{k}{m}}[/itex] where [itex]\omega[/itex] is the (angular) frequency of oscillation. The only way I know to do this is to solve the differential equation [itex]\ddot{y} + \frac{k}{m}y = 0[/itex] using [itex]y = A\cos(\omega t + \phi)[/itex] which gives [itex]\omega = \sqrt{\frac{k}{m}}[/itex]. Where does the [itex]\geq[/itex] sign come from?

Thanks.

James
 
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If there are no other forces acting apart from gravity and the spring, then it should be equals, not greater than or equal to.
 
That's what I thought, but apparently not so. If anyone would like to have a look at the original question, follow the link and go to question 8 (a)

http://www.physics.usyd.edu.au/ugrad/jphys/jphys_webct/jp_exams/1901_exam_2002.pdf
 
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Exam tomorrow. If anyone has any clues I'd love to know. . .

Thanks.

James
 
jdstokes said:
That's what I thought, but apparently not so. If anyone would like to have a look at the original question, follow the link and go to question 8 (a)

http://www.physics.usyd.edu.au/ugrad/jphys/jphys_webct/jp_exams/1901_exam_2002.pdf

The [tex]\ge[/tex] might apply if the spring were not linear, but the problem states that the motion is simple harmonic, which, by definition, means that the spring is linear for all of the motion. I assume the statement about ignoring gravity is there so that you will not be thrown off worrying about it, but gravity does not affect the frequency of the oscillator. It only affects the equilibrium position.

You can imagine a scenario in which the mass is pulled down too far and the spring is compressed until the coils touch so that the mass bounces down suddenly instead of being pushed down by the spring force. That would increase the frequency, but that is not SHM. The way the problem is worded, only the equal sign is valid.
 
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