Simple harmonic motion

  • Thread starter chouZ
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  • #1
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A mass of 0.100 kg is supported in equilibrium by two identical springs of negligible mass having spring constants k= 0.050 N/m. In the equilibrium position, the springs make an angle of 30 degree with the horizontal and are 0.100 m in length. If the mass, m, is pulled down a distance of 0.020 m and released, find the period of the resulting oscillation.



MY ATTEMPT:
angular frequency= square root (2k/m)

frequence= (1/2pi)angular frequency

Period= 1/frequency

It seems to be too easy like that so am not sure am right...:uhh:
 

Answers and Replies

  • #2
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I agree, it's way too simple an analysis. Most of the restoring force provided by the springs is spent neutralizing the other, and a minor fraction producing restoring force along the Y axis.
 
  • #3
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The other way I did it...I don't need the length of the spring nor the distance the mass is pulled down; numerical values given in the problem....Whoever try this, did u have to use those values???
 
  • #4
961
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lets look at the spring length between equilibrium and after the vertical displacement,
initially 0.1, from sin 30 we know the original y displacement = .05, and so the amplitude will vary between .03 and .07. The real trick is trying to figure out the k to use as its not simply k, the spring constant, but proportionate to the amt of additional stretch that works in the y direction.

see https://www.physicsforums.com/showthread.php?t=159470&page=2
 

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