Maximum extension in spring connecting two masses.

[Satvik Pandey]

Homework Statement

In the figure shown below all surfaces are friction-less. Find the maximum extension in the spring(in meters) , if the blocks are initially at rest and the spring is initially in its natural length.

Details and Assumptions:

F=30N
k=700N/m
m=5kg

Homework Equations

The Attempt at a Solution

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Let the displacement of $2m$ be $x$ and that of $m$ be $y$

Applying work energy theorem on block 1 ($2m$) and on block 2 we get

$Fx-\frac { 1 }{ 2 } k{ (x+y) }^{ 2 }=\frac { 1 }{ 2 } (2m){ v }_{ 1 }^{ 2 }$...(1)

$3Fy-\frac { 1 }{ 2 } k{ (x+y) }^{ 2 }=\frac { 1 }{ 2 } (m){ v }_{ 2 }^{ 2 }$....(2)

${ X }_{ COM\quad }=\frac { ml }{ 3m }$ Position of CoM initially.

${ X }_{ COM }=\frac { -2mx+m(L+y) }{ 3m }$ Position of CoM finally.

$\triangle { X }_{ COM }=\frac { my-2mx }{ 3m }$

${ a }_{ com }=\frac { 2F }{ 3m }$

${ v }_{ com }=\sqrt { 2\frac { 2F }{ 3m } \times \frac { my-2mx }{ 3m } }$.....(3)

Elongation in the string will be maximum when velocity of block 1 would be zero.

Putting this in (1) I got

$3x=35(x+y)$...(4)

On solving eq(2) and substituting eq(4) I got

$V_{2}=\sqrt(36y-12x)$...(5)

As $m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})V_{com}$

So $3V_{com}=v_{2}$
Using eq(5) and (4) I got

$-y=3x$
Am I correct till here?

 

[Satvik Pandey]

I am going to solve the equations. But I think I have done a mistake.

 

[haruspex]

I don't understand your equations 1 and 2. Aren't you counting the energy in the spring twice, once for the distance F moves and once for the distance 3F moves? Just consider total energy.
What is the relationship between the velocities at maximum extension?

 

[Satvik Pandey]

I don't understand your equations 1 and 2. Aren't you counting the energy in the spring twice, once for the distance F moves and once for the distance 3F moves? Just consider total energy.

I separately applied work energy theorem for block1 and 2.

What is the relationship between the velocities at maximum extension?

At maximum compression velocities of the blocks along the line joining them should be equal but this is a case of maximum extension. I think the extension would be maximum when velocity of mass $2m$ is zero.

 

[haruspex]

I separately applied work energy theorem for block1 and 2.

I don't see how you can do that. You have counted the total spring energy in each, so if you were to add them together you'd have too much. And there's no obvious way to apportion the spring energy between them.

At maximum compression velocities of the blocks along the line joining them should be equal but this is a case of maximum extension.

So? Why wouldn't it be the same result for the velocities?

 

[Satvik Pandey]

I don't see how you can do that. You have counted the total spring energy in each, so if you were to add them together you'd have too much. And there's no obvious way to apportion the spring energy between them.

So? Why wouldn't it be the same result for the velocities?

Work done by all forces is equal to the change in the kinetic energy of the system. Taking block of mass $2m$ as a system. So

Work done by F is $+Fx$
Work done by spring $-\frac{k(x+y)^{2}}{2}$ -ve sign because force exerted by the spring is in opposite direction of displacement.

Change in kinetic energy is $mv_{1}^{2}$ as the initial velocity was zero. Equating then I got eq(1) similarly I got eq(2).

What am I doing wrong?

 

[AlephNumbers]

Work done by F is +Fx

And what is the work done by the force on the other block?

 

[AlephNumbers]

What is the relationship between the velocities at maximum extension?

At maximum compression velocities of the blocks along the line joining them should be equal but this is a case of maximum extension. I think the extension would be maximum when velocity of mass 2m 2m is zero.

Kinetic energy is ½mv², right? So, according to your statement, the blocks should not have any kinetic energy when the spring has reached maximum extension.

 

[Satvik Pandey]

And what is the work done by the force on the other block?

If consider the other block to be moving rightward then
Work done by all the forces on that block would be $+3Fy-\frac{k(x+y)^2}{2}$.

 

[Satvik Pandey]

Kinetic energy is ½mv², right? So, according to your statement, the blocks should not have any kinetic energy when the spring has reached maximum extension.

I was making a mistake. I got that. The extension would be maximum at the moment when the relative velocity of the blocks along the line joining them would be zero.

 

[haruspex]

Work done by all forces is equal to the change in the kinetic energy of the system. Taking block of mass $2m$ as a system. So

Work done by F is $+Fx$
Work done by spring $-\frac{k(x+y)^{2}}{2}$ -ve sign because force exerted by the spring is in opposite direction of displacement.

Change in kinetic energy is $mv_{1}^{2}$ as the initial velocity was zero. Equating then I got eq(1) similarly I got eq(2).

What am I doing wrong?

k(x+y)²/2 is the total work done on the spring. You have both forces doing that much work on the spring.

 

[Satvik Pandey]

k(x+y)²/2 is the total work done on the spring. You have both forces doing that much work on the spring.

Oh! I think I got it.

Let us consider block 1,block 2 and spring as a system.

So work done by all the forces is equal to the change in the kinetic and potential energy of the system.

$Fx+3Fy=\frac{k(x+y)^2}{2}+\frac{3mv^2}{2}$ (extension would be maximum if velocities of blocks are equal).

Is work done by spring on block 1 $-\int _{ 0 }^{ x }{ k(x+y)dx }$?

 

[haruspex]

Is work done by spring on block 1 $-\int _{ 0 }^{ x }{ k(x+y)dx }$ ?

I see no basis for claiming that, and I see no need to plug in any assumption about it.
I feel there must be an easier way, but the only way I have been able to solve it so far is by writing out the ODEs for each block's motion and solving. There are some simplifications that can be made, e.g. by writing z = x+y.

 

[AlephNumbers]

Why is it, exactly, that you cannot use F = -kx? Is it because of the masses? The masses don't diminish the forces being applied on them, they just decrease the acceleration, right?

 

[haruspex]

Why is it, exactly, that you cannot use F = -kx? Is it because of the masses? The masses don't diminish the forces being applied on them, they just decrease the acceleration, right?

Of course you can use F = -kx (in this case, tension = k(x+y)). Where has anyone said otherwise?

 

[AlephNumbers]

Of course you can use F = -kx (in this case, tension = k(x+y)). Where has anyone said otherwise?

Oh. Okay. I guess no one said otherwise. When you said ODE I thought everything I knew probably went out the window.

 

[ehild]

I see no basis for claiming that, and I see no need to plug in any assumption about it.
I feel there must be an easier way, but the only way I have been able to solve it so far is by writing out the ODEs for each block's motion and solving. There are some simplifications that can be made, e.g. by writing z = x+y.

I think, that is the easiest method. You can introduce new variables, the position of the CM and the difference between the position of the masses, u=y-x. The equations of the motion separate to two single variable ODE-s. The acceleration of the CM is constant, and the distance between the masses performs SHM between the relaxed length and maximum length of the spring.

 

[Satvik Pandey]

I revised the chapter of Center of Mass last night and I found that this can be solved even in a more easy way.

Let us consider two blocks and the spring as a system.

Net external force acting on the system is $2F$

So by Newton's second law acceleration of the CoM of the system

$a_{com}=\frac{2F}{3m}$Consider a reference frame which is moving with the CoM of the system as it is a non-inertial frame of reference so we have to include Pseudo force on the blocks.

Let the displacement of block of mass $m$ be $x$ towards right and that of 2m be $y$ towards left as seen from the CoM of the system.

Work done by all the external forces on the right block is $(3F-\frac{2F}{3})x$Work done by all the external forces on the left block is $(\frac{4F}{3}+F)y$

This work done is equal to the change in the potential and kinetic energy of the system. If the extension has to be maximum then velocities of the block wrt CoM of the system should be zero. So the final kinetic energy of the system is zero.
$(3F-\frac{2F}{3})x+(\frac{4F}{3}+F)y= \frac{k(x+y)^2}{2}$Solving this equation I got $x+y=0.2$ which is the right answer.

 

[haruspex]

I revised the chapter of Center of Mass last night and I found that this can be solved even in a more easy way.

Let us consider two blocks and the spring as a system.

Net external force acting on the system is $2F$

So by Newton's second law acceleration of the CoM of the system

$a_{com}=\frac{2F}{3m}$Consider a reference frame which is moving with the CoM of the system as it is a non-inertial frame of reference so we have to include Pseudo force on the blocks.

View attachment 82092

Let the displacement of block of mass $m$ be $x$ towards right and that of 2m be $y$ towards left as seen from the CoM of the system.

Work done by all the external forces on the right block is $(3F-\frac{2F}{3})x$Work done by all the external forces on the left block is $(\frac{4F}{3}+F)y$

This work done is equal to the change in the potential and kinetic energy of the system. If the extension has to be maximum then velocities of the block wrt CoM of the system should be zero. So the final kinetic energy of the system is zero.
$(3F-\frac{2F}{3})x+(\frac{4F}{3}+F)y= \frac{k(x+y)^2}{2}$Solving this equation I got $x+y=0.2$ which is the right answer.

Congratulations.

 

[Satvik Pandey]

Congratulations.

Thank you!