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Mass Spring System

  • Thread starter frozen7
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  • #1
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In a vertical mass spring system, spring with spring constant 250 N/m vibrates with an amplitude of 12cm when 0.38kg hangs from it.
What is the equation describing this motion as a function of time? ( assume the mass passes through the equilibrium point, towards the positive x(upward), at t = 0.110s)

I do it in this way:
Hooke`s Law: F = kx
(0.38)(9.8)= 250 N/m (x)
x = 0.015(m)

[tex]\omega[/tex] = [tex]\sqrt{k/m}[/tex]
[tex]\omega[/tex] = 25.65

x = Acos([tex]\omega[/tex]t + [tex]\phi[/tex])
0.015 = 0.12cos(25.65t + [tex]\phi[/tex])
[tex]\phi[/tex] = -1.38

So, the equation : x = 0.12cos(25.65t -1.38)

Is it correct?
 

Answers and Replies

  • #2
mukundpa
Homework Helper
524
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1. Here the force is not mg, Actually mg is balanced by initial streach of the spring when it come in equilibrium position after the mass is attached.

2. F represents the restoring force trying to bring the mass back in the equilibrium position.
 
  • #3
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mukundpa said:
2. F represents the restoring force trying to bring the mass back in the equilibrium position.
The spring extent because of mg. Then why not F = mg?

How should I find the phi?
 
  • #4
mukundpa
Homework Helper
524
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What is the resultant force when the syatem is in equilibrium?

Actually the force F in S.H.M. is the restoring force on the mass when it has displacement x from equilibrium position.

your equation
[tex] x = A cos (\omega t + \phi ) [/tex]
is correct.

When x = 0 ; t = 0.110s
put these values to get [tex] \phi [/tex]
 
  • #5
164
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Ya, I did it in this way before also but the value I get is [tex]\phi[/tex] equal to -1.25

But it is actually should be about +1.89 ( from given answer)
 
  • #6
mukundpa
Homework Helper
524
3
[tex] cos \theta = cos \alpha [/tex] has general solution
[tex] 2n \pi \pm \alpha [/tex]

which sign is to be taken?
where was the mass at t = 0 ?
 
  • #7
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when t = 0
x = A

Any more clues?
 
  • #8
mukundpa
Homework Helper
524
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The same solution can be written as 3pi/2 - wt = 4.712 - 2.821 = 1.891 this is given in your text book.
 
Last edited:
  • #9
mukundpa
Homework Helper
524
3
The solutions of the equation are
[tex] \pi /2 - 2.821[/tex] and [tex] 3 \pi /2 - 2.821 [/tex]
= -1.25 and 1.891 radians respectively.
Now think, when t=0.11s; x=0 means at t = 0 the particle is yet to reach the equilibrium position [Where the phase angle (wt + phi) is pi/2] and x is - ve because the time period is 0.225s and 0.11s this is a bit less then half time period. For that the phase angle should be more then pi which is given by the value of phi = 1.891.
 
  • #10
164
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OK...Thanks...:)
 

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