Simple Harmonic Motion: Amplitude of Oscillation for a Spring System

AI Thread Summary
The discussion focuses on determining the amplitude of oscillation for mass m1 in a spring system after it loses contact with mass m2. The equations of motion for both masses are established, with emphasis on the normal contact force (N) becoming zero at the point of separation. Participants clarify that the motion of the center of mass (CM) can simplify the analysis, but it is not strictly necessary for solving the problem. The correct approach involves solving the second-order differential equation for displacement until separation occurs, after which a new equation applies for m1 alone. The conversation emphasizes the importance of understanding the dynamics of the system before and after the blocks separate.
Tanya Sharma
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Homework Statement



The masses in figure slide on a frictionless table.m1 ,but not m2 ,is fastened to the spring.If now m1 and m2 are pushed to the left,so that the spring is compressed a distance d,what will be the amplitude of the oscillation of m1 after the spring system is released ?

Homework Equations


The Attempt at a Solution

I feel if I can find the position and velocity where m1 loses contact with m2 ,i.e N=0 ,then new amplitude of oscillation can be obtained .

Let the origin be at m1 when the spring is unstretched.

The equation of motion for mass m1 is -kx-N = m1d2x/dt2 ,where N is the Normal contact force between the blocks and x is the displacement of the block.

I would be grateful if someone could help me with the problem.
 

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When a mass is oscillating on a spring, in some positions the spring is pushing, and in some it is pulling. Obviously, as soon as the spring starts pulling, the masses must separate.
 
voko said:
When a mass is oscillating on a spring, in some positions the spring is pushing, and in some it is pulling. Obviously, as soon as the spring starts pulling, the masses must separate.

Does that mean the blocks separate when the spring acquires its natural length ?
 
You explain what that means :)
 
How can i get the result ( i.e when the blocks separate ) mathematically ?

-kx-N = m1d2x/dt2

-kx-N = m1vdv/dx

Integrating,I get m1v2/2 = (k/2)(x+d)(d-x-N)

Should I put N=0 here ?
 
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I do not think you need to derive anything mathematically here, because the answer is obvious.

If you must, first consider the motion of the center of mass of the blocks. It is affected only by the external spring force.

Then, once you know this motion (and you also know that before separation the blocks are in contact), you can have equations for each block, which will have their interreaction force. It should be zero at the separation point, and with that you can locate the separation point.
 
EOM of CM : -kx=(m1+m2)d2x/dt2

EOM of m1 : -kx-N = m1d2x/dt2

EOM of m2 : N = m2d2x/dt2


Are these equations correct ? If yes , what should I do next ?
 
Before and just at the separation, x is the same in all three equations. At separation, N = 0. Solve for x.
 
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voko said:
Before and just at the separation, x is the same in all three equations. At separation, N = 0. Solve for x.

Could you please give your response to post#5 .
 
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  • #10
#5 is wrong. N is not constant, so you cannot integrate like that.

But from the equations in #7, you can find N as a function of x if you want to.
 
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  • #11
voko said:
#5 is wrong. N is not constant, so you cannot integrate like that.

Right...Thanks !

voko said:
But from the equations in #7, you can find N as a function of x if you want to.

Okay...so using EOM for m1 and m2 ,N=-kxm2/(m1+m2)

Now putting N=0 → x=0

Is this correct ?
 
  • #12
To begin with, both masses will be moving together. At what point in time will the contact force be equal to zero. After that, m1 will slow down faster than if they had been stuck together, while m2 will continue with its existing velocity. How does the motion of m1 vary after that time?
 
  • #13
Yes, x = 0 is correct.
 
  • #14
Chestermiller said:
At what point in time will the contact force be equal to zero.

Hi Chet :smile:

How should I find the time at which the contact force goes to zero ? Which equation should I use ?
 
  • #15
voko said:
If you must, first consider the motion of the center of mass of the blocks. It is affected only by the external spring force.

What is the significance of considering the motion of CM ?
 
  • #16
Tanya Sharma said:
Hi Chet :smile:

How should I find the time at which the contact force goes to zero ? Which equation should I use ?
The contact force goes to zero the first time that d2x/dt2 goes to zero. After that, m1 and m2 separate.
 
  • #17
Chestermiller said:
The contact force goes to zero the first time that d2x/dt2 goes to zero. After that, m1 and m2 separate.

So,i need to solve m2d2x/dt2 = 0 .

Right ?
 
  • #18
Tanya Sharma said:
What is the significance of considering the motion of CM ?

I think you obtained the equation for the inter-block force from considering the CM motion and the motion of one block. That should answer your question.
 
  • #19
voko said:
I think you obtained the equation for the inter-block force from considering the CM motion and the motion of one block. That should answer your question.

No...

I considered EOM of m1 and m2 .Knowledge of CM motion was not required .
 
  • #20
Tanya Sharma said:
No...

I considered EOM of m1 and m2 .Knowledge of CM motion was not required .

Ah, OK. The equation for CM is not strictly required, but I was not sure how you were going to solve the problem. It seemed to me that you wanted to obtain an explicit solution for x, which is very simple if you consider the CM.
 
  • #21
Tanya Sharma said:
So,i need to solve m2d2x/dt2 = 0 .

Right ?
You already showed that that happens the first time x = 0.
 
  • #22
Chestermiller said:
You already showed that that happens the first time x = 0.

Okay...I thought you were referring to some alternative approach by finding the time when N goes to 0.

I really like the way you approach the problems.Very neat mathematical ways .No guessing games :smile:
 
  • #23
voko said:
Ah, OK. The equation for CM is not strictly required, but I was not sure how you were going to solve the problem. It seemed to me that you wanted to obtain an explicit solution for x, which is very simple if you consider the CM.

If I need a solution for displacement(x) as a function of time(t) then I need to solve the second order DE d2x/dt2+[k/(m1+m2)]x = 0

Is it correct?
 
  • #24
Tanya Sharma said:
If I need a solution for displacement(x) as a function of time(t) then I need to solve the second order DE d2x/dt2+[k/(m1+m2)]x = 0

Is it correct?

Yes.
 
  • #25
voko...Thank you very much for your valuable guidance.
 
  • #26
Tanya Sharma said:
If I need a solution for displacement(x) as a function of time(t) then I need to solve the second order DE d2x/dt2+[k/(m1+m2)]x = 0

Is it correct?
Yes, but only up to the time that the two masses separate. After that, a new problem starts with
d2x/dt2+[k/m1]x = 0
 
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  • #27
Chestermiller said:
Yes, but only up to the time that the two masses separate. After that, a new problem starts with
d2x/dt2+[k/m1]x = 0

Yes..that's right . Excellent input.

Thanks !
 

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