Angular frequency of oscillation

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  • #1
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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  • #2
James R
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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.
 
  • #3
jdstokes
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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 [Broken]
 
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  • #4
jdstokes
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Exam tomorrow. If anyone has any clues I'd love to know. . .

Thanks.

James
 
  • #5
OlderDan
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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 [Broken]

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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