Simple Harmonic Motion: Amplitude of Oscillation for a Spring System

In summary: 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 use the equation for the contact force between the blocks. Can you tell me more about that?
  • #1
Tanya Sharma
1,540
135

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.
 

Attachments

  • spring.PNG
    spring.PNG
    355 bytes · Views: 343
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
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 ?
 
  • #4
You explain what that means :)
 
  • #5
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 ?
 
Last edited:
  • #6
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.
 
  • #7
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 ?
 
  • #8
Before and just at the separation, x is the same in all three equations. At separation, N = 0. Solve for x.
 
  • Like
Likes 1 person
  • #9
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 .
 
Last edited:
  • #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.
 
  • Like
Likes 1 person
  • #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
 
  • Like
Likes 1 person
  • #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 !
 

Related to Simple Harmonic Motion: Amplitude of Oscillation for a Spring System

1. What is simple harmonic motion?

Simple harmonic motion refers to the repetitive back-and-forth motion of an object that is caused by a restoring force or torque that is directly proportional to the displacement from equilibrium. This type of motion is commonly found in systems such as springs, pendulums, and mass-spring systems.

2. What is the amplitude of oscillation in a spring system?

The amplitude of oscillation in a spring system is the maximum displacement of the spring from its equilibrium position. It is represented by the distance between the equilibrium position and the highest or lowest point of the oscillation. In other words, it is the total distance the spring travels during one cycle of its motion.

3. How is the amplitude of oscillation related to the spring constant?

The amplitude of oscillation is directly proportional to the square root of the spring constant. This means that as the spring constant increases, the amplitude of oscillation also increases. Similarly, if the spring constant decreases, the amplitude of oscillation decreases as well.

4. Does the mass of the object affect the amplitude of oscillation in a spring system?

Yes, the mass of the object attached to the spring can affect the amplitude of oscillation. The larger the mass, the lower the amplitude of oscillation will be. This is because a heavier object requires more force to move it, resulting in a smaller displacement from equilibrium.

5. How does changing the initial displacement affect the amplitude of oscillation in a spring system?

The initial displacement of the object attached to the spring does not affect the amplitude of oscillation. The amplitude is solely determined by the properties of the spring, such as its spring constant and the mass of the object. However, the initial displacement will affect the position of the object at any given time during the oscillation.

Similar threads

What is the max amplitude before the sand particles slip off the vibrating plate?
    • Introductory Physics Homework Help
Replies
13
Views
339
Position and acceleration in simple harmonic motion given velocity
    • Introductory Physics Homework Help
Replies
16
Views
426
Newton's Second Law (Pulley Sustem)
    • Introductory Physics Homework Help
Replies
3
Views
828
Find the point of separation in SHM
    • Introductory Physics Homework Help
Replies
15
Views
1K
Does amplitude depend on mass in SHM?
    • Introductory Physics Homework Help
Replies
1
Views
6K
Why is my solution to finding the spring constant incorrect?
    • Introductory Physics Homework Help
Replies
3
Views
1K
Derive the formula for the frequency of a spring
    • Introductory Physics Homework Help
Replies
3
Views
1K
Kleppner - blocks and a compressed spring
    • Introductory Physics Homework Help
Replies
3
Views
2K
Zero Amplitude Damped Simple Harmonic Motion with k=0.7s^-1 and f=3Hz
    • Introductory Physics Homework Help
Replies
5
Views
861
Finding the amplitude of a vertical spring
    • Introductory Physics Homework Help
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top