Simple harmonic motion of an ideal spring

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  • #1
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simple harmonic motion !!!!!!!!!!!!!

Homework Statement


An ideal spring has a spring constant k=25N/M and is suspended vertically. A 1kg object is attatched to the unstrechted spring and released. It then performs ocillations.
What is the magnitude of the acceleration of the body when the extension of the spring is at maximum?
What is the maximum extension of the spring?




Homework Equations


x=Asin(wt)
v=Awcos(wt)=0
a=-Aw2sin(wt)
F=-kx-mg=ma.... im not sure why kx and mg are in the same direction on this....
a=-(w2)x



The Attempt at a Solution

im totally confused how to even start with this problem at this point.Can i use the fact that the force of a spring on a mass =-kx=ma or would that be incorrect would i need to use F=ma=-kx-mg? if i use the second one it really doesnt get me anywheres since i dont have much information to begin with..
 
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Answers and Replies

  • #2
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Well for a start, you have tons of information to begin with.

Now, the equation F = -kx, the minus sign is used to indicate that the force is a restoring force. Meaning, the force will always be directed to bring the mass back to equilibrium.
So, you have to set your y-direction properly. Either you want upwards as positive (which doesn't make sense in this case) or downwards as positive. Once you have done that, your will notice, the equation will start to make sense.
 
  • #3
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so your saying the equation should be F=-kx+mg=ma???? and why wouldnt it make sense to make up positive?
 
  • #4
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Well, i guess i went to far as to say i doesn't make sense. It is really up to preference, i would prefer my starting point to be x=0, and as the mass drops down x = positive number.

So, if you want x=negative number, that completely fine too.
 
  • #5
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ok well now im confused about the correct way to write the force equation...
should i include mg e.g. F=mg-kx=ma? or do i not include mg F=-kx=ma? i used the force equation earlier F=-kx=ma plugged in -Aw2sin(wt) for a....and Asin(wt) for x then solved for w and got w=5rad/s. This i just found out was the correct answer for w since i just found another equation T=2pi(m/k)^1/2 so my question is what force equation is the correct equation? Do you generally ignore gravity in harmonic motion problems?
 
  • #6
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Alright. Now, the confusion firstly arise because your equations are wrong to start with. They do not describe the situation in this example. (i.e. the initial conditions are different, you cannot just blindly use the equation your instructor gave)

Firstly, think about this. I want you to tell me which direction you are taking as positive, where is your x=0. Then i want you to give me the equation for x as a function of time. Make sure that your x equation describe the motion, and then differentiate and work from there.

It might also be helpful to draw a force vs x graph.

delzac
 
  • #7
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i dont think your correct telling me my equations are wrong or that they do not describe the situation in this example. x=Asin(wt) x is not equal to zero in this equation but is equal to A. v=Awcos(wt)=0 at max ext therefore wt= pi/2 or 3pi/2 which in turn means that t=0 at equillibrium.
 
  • #8
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Think about this:

Firstly, like what i said, where do you take x =0? In all likelihood you will choose the point where the spring is unstretched, when the mass is first attached, but not released.
Therefore when t= 0, does your equation give you x=0? If you are using the equation " x=Asin(wt) x ", it does give x=0. So far so good.

Then you check again.

I told you to state your positive direction. Which do you take as up or down? Regardless. You notice that your equation " x=Asin(wt) x" for different values of t, will give positive and negative x. Does it make sense to get a positive and negative x for different t in relation to this example?

So, you have to define where is x=0, where is your positive and negative direction and you work from there. Your x equation will have to FOLLOW the initial conditions, because in every question, something will be different and you have to adjust your equation accordingly.
 
  • #9
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Ok well if I change the equation to x=Acos(wt) then at t =0. Cos(wt)=1
And cos(wt) still gives both pos and neg values for x..
 
  • #10
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Oh and down will be positive
 
  • #11
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For a start, you should drawn out a x vs t graph, then you will know what shape the curve is and the equation will automatically follow.
 
  • #12
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Ok well lookin at it that way I would say that Acos(wt) would be the correct one to use since when x=0 the graph is at its max amplitude
 
  • #13
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Thats correct. But you know that just saying x = Acos(wt) is not enough. Something is still off. What is off? And how to you adjust it into your equation. Drawn out that graph, and adjust your equation accordingly.
 
  • #14
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Need to move the graph up one. I forgot how to do that... Is it Acos(wt)-1?
 
  • #15
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Yes, you will need to shift the curve up. How do you shift your straight line up the graph?

Also, is the curve Acos(wt) or -Acos(wt)?

And how much do you shift? Are you sure it is by 1? Have you marked out the axises on your graph? Meaning where the A in the y= A(sinusoidal function) is on the axis?

After you have done that, continue to differentiate to obtain Velocity and acceleration.

From acceleration you will know what you force is simple because F=ma. But, take note. That force is not spring force, it is net force that is on the mass.
 
  • #16
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Ok is this correct -Acos(wt)+A. Neg so when x is zero t is zero and up A so that the graph is always pos
 
  • #17
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Great! Now differential twice and get acceleration, find your net force (spring force and weight) acting on the mass. Then try solving for w.
 
  • #18
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Ok I get v=Awsin(wt) and a=Aw^2cos(wt). Then I plug in the x equation and a equation into F=mg-kx=ma and I get g=Aw^2
 
  • #19
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Now you notice that w^2 = g/A. It would look like it is dependent on gravity. But it is not (you can infer this since A will increase as gravity increases, meaning there is a g component in A). You can convert g/A to something else. Think energy.
 
  • #20
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I'm thinking A could represent max potential energy 1/2kx^2
 
  • #21
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A has the dimensions of metre, it can't possibly have units of Joules.

Try and equate total energy at the top and bottom and see what you get.
 
  • #22
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I get A=x but I already knew that.. g/x. Maybe the work done by gravity?
 
  • #23
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Nope. F = -kx or E = 1/2 kx^2, the x represents the extension of the spring. It is a variable, and it is constantly changing. Your A is a fixed value.
 
  • #24
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Well I think at the top the total energy is zero and at the bottom it is 1/2kx^2. According to the equation I have for E=1/2mv^2+1/2kx^2
 
  • #25
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If the total energy at the top is zero, and total energy at bottom is a finite value, then does it mean that conservation of energy is not conserved? Surely it is not the case. At the x=0, v=0. At x=max, v=0. Thus at top and bottom there is no kinetic energy. So what are there? Equate them and try to get g/A.
 

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