Simple Harmonic Motion of a spring and a mass

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  • #1
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Homework Statement



A spiral spring of natural length 300mm is suspended vertically from a fixesd point with ita upper end fixed .A mass of 0.150 kg is suspended at rest from the lower end of the spring to increase its length to 355mm.The mass is then pulled down a further distance of 30 mm and released from rest so it osciallates about equlibrium.
Calculate
*spring constant k
*time period of oscillations.
*the max speed and max ke of the mass.
*max and minimum tension in the spring.

Homework Equations


k=fx, time= 2pie/omega,k.e=1/2mv2


The Attempt at a Solution


* for the first bit i got k =0.150x0.085=0.01275Nm/s (not sure)
*how to find time period, as T=2pie/omega or 1/f =( hm..am i missing something here
 

Answers and Replies

  • #2
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OO i think i got it, i have to use T=2pie[tex]\sqrt{}k/m[/tex]?
if thats the case.. then how do i get max speed for the 3rd bit .. as i dont have amplitude.. and i cant use max speed= A(omega), i am confused as this formula is for max velocity..=/ (or do i simple use v= distance/ time?
 
  • #3
Redbelly98
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Homework Equations


k=fx, time= 2pie/omega,k.e=1/2mv2


The Attempt at a Solution


* for the first bit i got k =0.150x0.085=0.01275Nm/s (not sure)
*how to find time period, as T=2pie/omega or 1/f =( hm..am i missing something here

For the first part, look up the Hooke's Law equation again. It's not "k=fx", but is similar to that. You also need to use the weight of the 0.150 kg mass ... and the displacement is not 0.085 m.

For the 2nd question, you're correct that T=2π/ω. If you get ω in terms of k and m, you should be able to work it out then.
 
  • #4
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ah thanks for the reply. oo yea. its f= -kx , sorry need to concentrate while typing , so if its not 0.085 displacement.. hm then what is it =/
 
  • #5
Redbelly98
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Well, the spring was originally 300 mm. Hanging the mass on the spring and letting it come to rest, the new spring length is 355 mm. So the change in length due to the added weight is ____?
 
  • #6
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ah sorry been away for sometime ermm.. the change would be 55mm?>but in the question it says.. the spring is further extended by 30 mm ..so thats why i thought total extension is 85mm =/.
 
  • #7
Redbelly98
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When it is pulled to 85 mm, the force is given by the weight PLUS the force needed to pull it the extra 30 mm. Since we don't know how much that extra force is, we can't plug in for F in F=kx.

When it is just the weight of the mass pulling against the spring, the change in length is 55 mm.
 
  • #8
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ah cool.. thanks alot =]
 

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