Spring mechanics

  • #1
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Homework Statement



F <-- M------------------------------------M →F


Two equal masses M are connected by a spring having spring constant k, as shown above.
We have to find the maximum elongation of the spring if two equal forces F are applied in opposite directions..

Homework Equations



Please, there are many equations.

The Attempt at a Solution



I solved it via two methods and those two give different answers.

Due to symmetry the centre of mass will be at mid point and let elongation at either side be x, Now the spring constant of each half will be 2k.

First method: Work done by the forces on a system = Change in kinetic energy + Change in potential energy of the system :

2Fx = 0 + 2(2kx2)/2
thus x = F/k
Therefore total elongation = 2x = 2F/k

Method II :

F = 2kx
or x = F/2k
Total elongation = 2x = F/k


Which answer is wrong ?

Please help !!
 

Answers and Replies

  • #2
TSny
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Method II :
F = 2kx
or x = F/2k
Total elongation = 2x = F/k
At the instant of time when F = 2kx, what can you say about the acceleration of each mass? What can you say about the velocity of each mass?
 
  • #3
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At the instant of time when F = 2kx, what can you say about the acceleration of each mass? What can you say about the velocity of each mass?
The net force on each mass as a function of extension x is,

F(x) = F-2kx

When x = xmaximum
F= 2kxmaximum
Then,

F(xmaximum) = F - 2kxmaximum = 2kxmaximum-2kxmaximum=0 , that is zero net force on each mass.

Also, before performing simple harmonic motion, there is no kinetic energy on each mass. Hence zero velocity on each mass. Correct ?
 
  • #4
TSny
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Yes, the net force on the mass on the right is Fnet = F - 2kx where x is the amount the mass has moved. At the beginning, x = 0 and the net force on the mass is positive. So the mass has positive acceleration at the beginning and therefore has increasing speed.

As the spring stretches the net force decreases. At what point is the net force zero? What is the acceleration of the mass at that point? Is the velocity zero at that point? Has the mass reached it's maximum value of x at this point?
 
  • #5
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Yes, the net force on the mass on the right is Fnet = F - 2kx where x is the amount the mass has moved. At the beginning, x = 0 and the net force on the mass is positive. So the mass has positive acceleration at the beginning and therefore has increasing speed.
We are just applying force in a negative potential gradient : -kx. There will only be potential energy and not kinetic energy..

As the spring stretches the net force decreases. At what point is the net force zero? What is the acceleration of the mass at that point? Is the velocity zero at that point? Has the mass reached it's maximum value of x at this point?
At the amplitude, that is at the maximum extension will the net force on each mass be zero. There will be no acceleration, also no velocity. I can see that method 1, that is conservation of energy, gave, maximum extension= 2F/k, but method 2 gave F/k. Why such a contradictory answer in both the logical solutions ?
 
  • #6
TSny
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We are just applying force in a negative potential gradient : -kx. There will only be potential energy and not kinetic energy.
The masses will have kinetic energy as they move to their maximum extension. Only at the point of maximum extension will their kinetic energy be zero.

At the amplitude, that is at the maximum extension will the net force on each mass be zero. There will be no acceleration, also no velocity. I can see that method 1, that is conservation of energy, gave, maximum extension= 2F/k, but method 2 gave F/k. Why such a contradictory answer in both the logical solutions ?
At the maximum extension, the net force will not be zero.

Here is a similar situation: Suppose you have a spring hanging vertically from a support. You then attach a mass on the end of the spring and let the mass go from rest from the position where the spring is unstretched. There will be two forces acting on the mass, the force of the spring and the constant force of gravity. The force of gravity is taking the place of the applied force F in your problem. As the mass descends, it will gain kinetic energy until the two forces are equal in magnitude. At that point the mass has maximum kinetic energy. After that, the upward spring force is greater than the downward force of gravity and the mass starts slowing down. When it reaches the point where the mass momentarily comes to rest, it should be clear that the spring force will be greater than the force of gravity.
 
  • #7
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The masses will have kinetic energy as they move to their maximum extension. Only at the point of maximum extension will their kinetic energy be zero.
TSny, before performing SHM, when we merely elongate the string there will be no kinetic energy, just the gain in potential energy of k(x1+x2)2/2 which is equal to the work done against force -kx, which is F(x1+x2) where x1 and x2 are maximum elongations of each of force F on either sides.

Hence F(x1+x2) = k(x1+x2)2/2

which yields,

x1+x2 = 2F/k...

Here two blocks are M1 and M2 and M1≠M2.

If M1=M2=M, say, then by concept of centre of mass of the system, x1=x2=x.

Then also we get, 2x=2F/k , using conservation of mechanical energy.

But if we apply symmetry in case M1=M2=M => x1=x2=x , we get 2x=F/k...

One book say its 2F/k, while other two books give F/k.
Note the book which give 2F/k as answer assumes mass of two blocks are M1 and M2 and M1≠M2, while the other two books assumes M1=M2=M. But merely changing the masses should not affect the maximum elongation of the spring.

I agree that while SHM, we cannot say net force be zero at amplitude.

At the maximum extension, the net force will not be zero.

Here is a similar situation: Suppose you have a spring hanging vertically from a support. You then attach a mass on the end of the spring and let the mass go from rest from the position where the spring is unstretched. There will be two forces acting on the mass, the force of the spring and the constant force of gravity. The force of gravity is taking the place of the applied force F in your problem. As the mass descends, it will gain kinetic energy until the two forces are equal in magnitude. At that point the mass has maximum kinetic energy. After that, the upward spring force is greater than the downward force of gravity and the mass starts slowing down. When it reaches the point where the mass momentarily comes to rest, it should be clear that the spring force will be greater than the force of gravity.
Also a side question :

Suppose a simple pendulum is in SHM. Then its equation is, y=A(sinωt). Then at y=A, ωt=90 degrees. Then this means that always the amplitude is when string of pendulum makes angle 90 degree with vertical. Then how can this be when pendulum can undergo SHM in small oscillations, where it can never reach the point where angle made by vertical is 90 degrees ?
 
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  • #8
Doc Al
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TSny, before performing SHM, when we merely elongate the string there will be no kinetic energy, just the gain in potential energy of k(x1+x2)2/2 which is equal to the work done against force -kx, which is F(x1+x2) where x1 and x2 are maximum elongations of each of force F on either sides.
Realize that when you stretch the string gradually, so as not to add kinetic energy, the force you must exert is not a constant. So the work done is not simply Fx, since F is now a function of x. ∫Fdx = 1/2kx2

In your original problem, presumably F is a constant force so the work done by it is Fx. That's where the additional energy, and thus elongation, comes from.

Also a side question :

Suppose a simple pendulum is in SHM. Then its equation is, y=A(sinωt). Then at y=A, ωt=90 degrees. Then this means that always the amplitude is when string of pendulum makes angle 90 degree with vertical.
ωt does not represent the angle that the pendulum makes with the vertical.
 
  • #9
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Realize that when you stretch the string gradually, so as not to add kinetic energy, the force you must exert is not a constant. So the work done is not simply Fx, since F is now a function of x. ∫Fdx = 1/2kx2

In your original problem, presumably F is a constant force so the work done by it is Fx. That's where the additional energy, and thus elongation, comes from.
Ok, I know that, just mistyped. By using symmetry the answer comes to be "maximum elongation"=F/k, and by using conservation of energy, in the "attempt at solution", maximum elongation= 2F/k.

God knows which one is correct !! :cry:

The two different answers were also given in the "three" different books.

As is stated:

Using conservation of energy :

Hence F(x1+x2) = k(x1+x2)2/2

which yields,

x1+x2 = 2F/k...

Here two blocks are M1 and M2 and M1≠M2.

If M1=M2=M, say, then by concept of centre of mass of the system, x1=x2=x.

Then also we get, 2x=2F/k , using conservation of mechanical energy.

But if we apply symmetry in case M1=M2=M => x1=x2=x , we get 2x=F/k...

One book say its 2F/k, while other two books give F/k.
Note the book which give 2F/k as answer assumes mass of two blocks are M1 and M2 and M1≠M2, while the other two books assumes M1=M2=M. But merely changing the masses should not affect the maximum elongation of the spring.
 
  • #10
Doc Al
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By using symmetry the answer comes to be "maximum elongation"=F/k,
What do you mean by "using symmetry"? Looks to me you tried to set F = kx, which only gives the point where the spring force equals the applied force, not the point of maximum elongation.

The two different answers were also given in the "three" different books.
What books? Are you sure the problems in each book are identical?
 
  • #11
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What do you mean by "using symmetry"? Looks to me you tried to set F = kx, which only gives the point where the spring force equals the applied force, not the point of maximum elongation.


What books? Are you sure the problems in each book are identical?
Ok, Doc. I hereby quote the solution of two books :

Book 1:

Problem Statement : Consider the following system:

F <-- M1------------------------------------M2 →F

Suppose each of the blocks having masses M1 and M2 are pulled by a constant force F applied, as shown above. Find the maximum elongation the spring (shown by ----) will suffer. Spring constant = K

Solution :
If both the blocks are pulled by some force, they suddenly move with some acceleration and
instantaneously stop at same position where the elongation of spring is maximum.
Let x1, x2 → extension by block M1 and M2
Total work done = Fx1 + Fx2…(1)
Increase the potential energy of spring = K (x1+ x2)2/2 …(2)
Equating (1) and (2)
F(x1 + x2) = K (x1+ x2)2/2 => (x1+ x2) = 2F/k

Hence maximum elongation = 2F/k

Book 2:

Problem statement : Consider :

F <-- M------------------------------------M →F

See the above figure. Spring constant =K. The two blocks attached have same mass and force F is applied as shown above. Find the maximum extension of the spring.

Solution : When forces F are applied on masses M in opposite directions as shown, the middle point "O" of the spring remains stationary at its position. Therefore length of the spring "l" (say) can be considered to be made up of two parts, each of length "l/2", and hence the spring constant 2k, and joined to each other at the point "O". Since the extension is proportional to the length of the spring, each part undergoes an extension by,

x/2 = F/2k

Total extension (maximum) in the spring is,

x=x/2 + x/2 = F/2k + F/2k = F/k

______________________________________________________________________

So both the answers are different. Also I do not think that its the mass of block which is affecting the answer.

Can you explain it Doc Al ? Thanks.
 
  • #12
Doc Al
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Book 1:

Problem Statement : Consider the following system:

F <-- M1------------------------------------M2 →F

Suppose each of the blocks having masses M1 and M2 are pulled by a constant force F applied, as shown above. Find the maximum elongation the spring (shown by ----) will suffer. Spring constant = K

Solution :
If both the blocks are pulled by some force, they suddenly move with some acceleration and
instantaneously stop at same position where the elongation of spring is maximum.
Let x1, x2 → extension by block M1 and M2
Total work done = Fx1 + Fx2…(1)
Increase the potential energy of spring = K (x1+ x2)2/2 …(2)
Equating (1) and (2)
F(x1 + x2) = K (x1+ x2)2/2 => (x1+ x2) = 2F/k

Hence maximum elongation = 2F/k
Perfectly clear and correct.


Book 2:

Problem statement : Consider :

F <-- M------------------------------------M →F

See the above figure. Spring constant =K. The two blocks attached have same mass and force F is applied as shown above. Find the maximum extension of the spring.

Solution : When forces F are applied on masses M in opposite directions as shown, the middle point "O" of the spring remains stationary at its position. Therefore length of the spring "l" (say) can be considered to be made up of two parts, each of length "l/2", and hence the spring constant 2k, and joined to each other at the point "O". Since the extension is proportional to the length of the spring, each part undergoes an extension by,

x/2 = F/2k

Total extension (maximum) in the spring is,

x=x/2 + x/2 = F/2k + F/2k = F/k
A completely different problem. Note how they apply Hooke's law to get the extension--that would only be true at equilibrium. This force F is not constant, as it was in the first problem.

(If they meant this to be the same as problem 1, then they are wrong. I suspect that they did mean it to be the same, since they ask about the "maximum" extension. At equilibrium, there is just an extension; "maximum" doesn't apply.)

You can certainly solve the first problem by dividing the spring into two parts. You'll get the same answer, of course.


So both the answers are different. Also I do not think that its the mass of block which is affecting the answer.
The books are addressing different problems, whether they intend to do so or not. The second problem is poorly worded if they just meant it to be an application of Hooke's law, as the solution suggests.
 
  • #13
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Perfectly clear and correct.



A completely different problem. Note how they apply Hooke's law to get the extension--that would only be true at equilibrium. This force F is not constant, as it was in the first problem.

(If they meant this to be the same as problem 1, then they are wrong. I suspect that they did mean it to be the same, since they ask about the "maximum" extension. At equilibrium, there is just an extension; "maximum" doesn't apply.)

You can certainly solve the first problem by dividing the spring into two parts. You'll get the same answer, of course.



The books are addressing different problems, whether they intend to do so or not. The second problem is poorly worded if they just meant it to be an application of Hooke's law, as the solution suggests.
After searching many books and looking on many many references, I came to the conclusion that solution of book 2 is not improper and not correct, whether they just mean to do it or not, as Doc Al says.

Thanks to both the posters here !! :smile:
 
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  • #14
orientor
Why will the moment of instantaneous rest be same when masses are different?
 
  • #15
TSny
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Welcome to PF!
Why will the moment of instantaneous rest be same when masses are different?
Suppose the masses are m1 and m2.

One way to get an answer to your question is to consider the following questions:
How does the net force acting on m1 compare to the net force acting on m2 at any instant of time?
How does the acceleration of m1 compare to the acceleration of m2 at any instant of time?
How does the velocity of m1 compare to the velocity of m2 at any instant of time?

Another approach is to use the fact that the velocity of the center of mass of a system cannot change if the net external force acting on the system is zero.
 
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  • #16
orientor
If we look at the forces, then net force on M1= F-kx
Net force on M2=F-kx

Hence net force will be same.

For acceleration,

Acceleration of M1- (F-kx)/M1

Acceleration of M2- (F-kx)/M2

Hence acceleration will be different.

Since net force is same but acceleration is different, therefore velocities will also be different.

So I think the moment of rest will be different for them.

Sorry for the bad formatting.
 
  • #17
TSny
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If we look at the forces, then net force on M1= F-kx
Net force on M2=F-kx

Hence net force will be same.
Yes, the magnitudes of the forces will be equal. Good. I assume that you are taking the symbol x to represent the total amount of stretch of the spring at the instant of time that you are considering. In earlier parts of this thread, x denoted half the total stretch.

For acceleration,

Acceleration of M1- (F-kx)/M1

Acceleration of M2- (F-kx)/M2

Hence acceleration will be different.
Yes, the accelerations will be different. But will they be proportional? That is, at any instant of time can you write a2 = c a1 where c is a constant? Can you express c in terms of the masses?

Since net force is same but acceleration is different, therefore velocities will also be different.
Yes, at an arbitrary time t, the speeds of the masses will generally be different.

So I think the moment of rest will be different for them.
To see if this is right, the question about the accelerations that I posed above will be helpful. A relation between the accelerations can be used to get a relation between the velocities. This will help to see if the velocity of both masses will be zero at the same time.
 
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  • #18
orientor
a1= (M2/M1) a2
So, v1=(M2/M1)v2.

So if v1 is zero then v2 will also be zero. So, their moment of instantaneous rest will be same.

Thanks for guiding me!!
 
  • #19
TSny
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a1= (M2/M1) a2
So, v1=(M2/M1)v2.

So if v1 is zero then v2 will also be zero. So, their moment of instantaneous rest will be same.

Thanks for guiding me!!
Good work!
 

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