Two objects attached with a spring

[giokrutoi]

yes I forgot that but I don't think that it's the write solution because that 10 centimeters are maximal extension for both but not for m2

 

[jbriggs444]

maximal compression may be 10 centimeters and if that's true force should be 10 centimeter multiplied by k and acceleration of second mass should be this magnitude divided by m2
and yes I mean right when I write write

That simple analysis is attractive, but incorrect. After m1 is released and before m2 is released, m1 is undergoing simple harmonic motion. It is easy to see that the maximal compression is 10 cm shorter than relaxed and that the maximal extension is 10 cm longer than relaxed. This motion can optionally be maintained for many cycles before m2 is finally released.

When m2 is released, the spring is at its relaxed length. If m2 were still held stationary, the spring would eventually extend 10 cm longer than relaxed. But because m2 has been released, we are no longer assured that this will be the case.

One line of analysis that indicates that it cannot be the case is based on energy. Think initially about the state with mass m2 pinned in place. In its compressed state, the spring has a certain amount of potential energy. That amount of energy can be calculated based on the 10 cm number and the spring constant. In its extended state, the spring has that same amount of potential energy. In both cases the system has zero kinetic energy. All of the energy is in the spring.

The pin holding m2 in place is doing no work. It applies a force, but it is motionless. Following release the pin holding m2 in place still does no work. Accordingly, energy is conserved within the m1, m2, spring system both before and after the release of m2.

If m1 and m2 were both motionless then the spring would have to extend or compress to the full 10 cm in order for energy to be conserved. But as long as either are in motion, the spring cannot ever extend that far without violating energy conservation. But momentum is conserved following release of m2, so it can never be the case that both m1 and m2 are motionless following the release of m2.

 

[giokrutoi]

That simple analysis is attractive, but incorrect. After m1 is released and before m2 is released, m1 is undergoing simple harmonic motion. It is easy to see that the maximal compression is 10 cm shorter than relaxed and that the maximal extension is 10 cm longer than relaxed. This motion can optionally be maintained for many cycles before m2 is finally released.

When m2 is released, the spring is at its relaxed length. If m2 were still held stationary, the spring would eventually extend 10 cm longer than relaxed. But because m2 has been released, we are no longer assured that this will be the case.

One line of analysis that indicates that it cannot be the case is based on energy. Think initially about the state with mass m2 pinned in place. In its compressed state, the spring has a certain amount of potential energy. That amount of energy can be calculated based on the 10 cm number and the spring constant. In its extended state, the spring has that same amount of potential energy. In both cases the system has zero kinetic energy. All of the energy is in the spring.

The pin holding m2 in place is doing no work. It applies a force, but it is motionless. Following release the pin holding m2 in place still does no work. Accordingly, energy is conserved within the m1, m2, spring system both before and after the release of m2.

If m1 and m2 were both motionless then the spring would have to extend or compress to the full 10 cm in order for energy to be conserved. But as long as either are in motion, the spring cannot ever extend that far without violating energy conservation. But momentum is conserved following release of m2, so it can never be the case that both m1 and m2 are motionless following the release of m2.

I understood what you have said and may guess was very silly but how can I solve this problem i wrote conservation of momentum up in commentary's and some of them said it was wrong
so please if you can read my upper comments and tell me were my conclusions wrong or not

 

[SammyS]

yes I forgot that but I don't think that it's the write solution because that 10 centimeters are maximal extension for both but not for m2

Do you mean "right" when you write "write"?

I was going to point this out, but ehild has done so previously.

"Right" = Correct ≠ "Write"

Right ?​

 

[giokrutoi]

I was going to point this out, but ehild has done so previously.

"Right" = Correct ≠ "Write"

Right ?​

ok I made mistake sorry

 

[giokrutoi]

I was going to point this out, but ehild has done so previously.

"Right" = Correct ≠ "Write"

Right ?​

but what can you say about solute

 

[SammyS]

but what can you say about solute

Too many helpers at one time is not good.

ehild and briggs are excellent.

 

[giokrutoi]

Too many helpers at one time is not good.

ehild and briggs are excellent.

but whom should I trust they are saying opposite things

 

[jbriggs444]

but whom should I trust they are saying opposite things

We are saying identical things.

 

[giokrutoi]

We are saying identical things.

but you say that it is impossible 10cm to be maximal extention

 

[jbriggs444]

but you say that it is impossible 10cm to be maximal extention

I say that it is not possible for 10 cm to be the maximal extension following release of m2. I do not see ehild making any contrary claim.

Edit: In any case, one should judge the truth of the matter by the strength of the arguments offered, not based on trust. If you can follow the arguments and see that the results follow, all is well. .

 

[giokrutoi]

this

 

[jbriggs444]

this

That says nothing about 10 cm.

 

[giokrutoi]

in attached file he says that max compression is 10 cm and thing is to convert it to meters

 

[giokrutoi]

he doesn't say he agrees
sorry

 

[giokrutoi]

?

 

[jbriggs444]

Rather than posting attachments which are screen shots of previous postings in this same thread, the preferred practice is to reference a previous message by number.

For instance, "see ehild's response in #50"

Or, even better, go back to response #50, mouse over the quotation, hit the "Quote" button, come back here and "Insert Quotes" like so:

Yes, but you need to calculate the compression in meters, and that should be multiplied by k=20 N/m.

Edit: And rather than replying multiple times in sequence, consider going back to your previous post and adding content. Like I just did here.

 

[giokrutoi]

see you are saying opposing things

 

[jbriggs444]

see you are saying opposing things

No. We are not saying opposing things.

Edit: Arguing about who said what does not advance anyone's understanding.

Is 10 cm the maximum extension after release of m2? I say no. If you say yes, advance your argument. Do not say "because ehild said so".

 

[giokrutoi]

can you explain that

as I understood ehild says that only thing I have to do is to convert 10cm to meters
you say that maximal compression is not 10cm
ok you convinced me I didn't pay attention on that

Sorry

 

[giokrutoi]

I say that it is not possible for 10 cm to be the maximal extension following release of m2. I do not see ehild making any contrary claim.

Edit: In any case, one should judge the truth of the matter by the strength of the arguments offered, not based on trust. If you can follow the arguments and see that the results follow, all is well. .

I think that your argument is more convincing

 

[jbriggs444]

I dislike playing he said she said. But...

maximal compression may be 10 centimeters and if that's true force should be 10 centimeter multiplied by k and acceleration of second mass should be this magnitude divided by m2
and yes I mean right when I write write

Yes, but you need to calculate the compression in meters, and that should be multiplied by k=20 N/m.

That's ehild agreeing that you have to multiply the maximal compression by k=20 N/m but NOT agreeing that the maximal compression is 10 cm.

Edit: Now with that said, how would one go about calculating the maximal compression following the release of m2? There are at least two general approaches that could work.

One approach would be to try to figure out how the new simple harmonic motion would look following the release of m2. Pick a fixed point (the center of mass?) and try to describe the motion of each piece around that point.

Another approach would be to consider energy. How much kinetic energy is there in the bulk motion of the combined system containing m1, m2 and the spring? How much is in their individual motions relative to the center of mass?

 

[ehild]

maximal compression may be 10 centimeters and if that's true force should be 10 centimeter multiplied by k and acceleration of second mass should be this magnitude divided by m2

Sorry, I made a mistake, the maximum distortion is not 0.1 m when both blocks move. You have to determine it first.

I suggest to follow jbriggs' hint concerning energy. You were about on the right track. But you have to define the notations you use.
After the system has been released, the center of mass moves with constant velocity (as there is no external force any more) and the spring with the attached masses performs simple harmonic motion (it vibrates) with respect to the center of mass. The kinetic energies of these two modes of motion add up. You get the energy of the vibration by subtracting the translational energy from the whole kinetic energy the system has when the second block is released and the spring is relaxed. That kinetic energy is converted to maximum potential energy when the distortion of the spring is maximum. But the kinetic energy of the vibration is not 1/2 m₂ x₂². What is x₂? and the vibration involves both masses.

 

[giokrutoi]

i was thinking about this problem and i came to conclusion that post 33 makes sense if v2 is maximal

EDIT: If that's wrong than period of second object maybe 2 pi m2/k₂ ^1/2 were k₂ is different from k which is given because according to yang's theory when spring's length is decreased (it is decreased because I write origin of spring in the center of mass) then k must go up so if I call Z as length of spring when it is relaxed then if I subtract length from m1 to center of mass I will get length of spring from center of mass to m2 and with help of proportion I will calculate k2 which will be higher that original k
can this help me?

 

[giokrutoi]

?

 

[BvU]

I have stated that before

Well, the 20 N/m is new to me. May I remind you of post # 2? You are now at over 70 posts and still drag in new things (yang ??) that indicate to me that post #6 still applies.

I am surprised the thread drags on so long; I apparently missed the last notification mail and wondered how much further you were.

You have had guidance from three excellent helpers (plus me , and a comment from yet another excellent member) -- by the way (post 65), ehild is a she -- and are spending way too much energy on what should be a fairly simple exercise.

Pick up some of the advice (work more systematic, be clear about notation and symbol meaning, slow down a bit ) and try the energy approach as suggested in 72 and 73.

Another suggestion: you do this exercise in a certain context in the curriculum material. Most exercises are constructed to let you practice with the lecture material in the current section, plus some basic general stuff and math. If yang were really needed you would have gotten a hint in the problem statement. Elizabeth's excellent (I don't know russian, but fully trust that 43 == 39 ) translation hasn't got a single indication.

 

[ehild]

i was thinking about this problem and i came to conclusion that post 33 makes sense if v2 is maximal

EDIT: If that's wrong than period of second object maybe 2 pi m2/k₂ ^1/2 were k₂ is different from k which is given because according to yang's theory when spring's length is decreased (it is decreased because I write origin of spring in the center of mass) then k must go up so if I call Z as length of spring when it is relaxed then if I subtract length from m1 to center of mass I will get length of spring from center of mass to m2 and with help of proportion I will calculate k2 which will be higher that original k
can this help me?

What is v2? what do you mean that it is maximal?

If you take the spring as two strings in series, connected at the centre of mass, you can determine the spring constant of both, and also the vibration frequency. Go ahead. But I do not see how you want to proceed and get the maximum acceleration of block 2.

 

[giokrutoi]

What is v2? what do you mean that it is maximal?

If you take the spring as two strings, connected at the centre of mass, you can determine the spring constant of both, and also the vibration frequency. Go ahead.

in file which i have attached to post 33 there is v2 and I state that it is maximum
but i guess that conclusion( in post 33) is wrong because i didn't take in consideration that all the energy can't be in m2

now period of second mass is 2 pi square root from m2 / k2 k2 is spring constant derived from here
if length of spring is Z then center of mass must be on that spring but it must be closer to the bigger mass so if i call distance from m2 to center of mass D than I have to find D
D is derived from here m1(Z-D)=m2D D=Z/m1+m2 so if k constant from Z is k then for D length it is k(m1+m2)
so period is 2 p square root from m2/k(m1+m2)
is that right

 

[ehild]

in file which i have attached to post 33 there is v2 and I state that it is maximum
but i guess that conclusion( in post 33) is wrong because i didn't take in consideration that all the energy can't be in m2

now period of second mass is 2 pi square root from m2 / k2 k2 is spring constant derived from here
if length of spring is Z then center of mass must be on that spring but it must be closer to the bigger mass so if i call distance from m2 to center of mass D than I have to find D
D is derived from here m1(Z-D)=m2D D=Z/m1+m2 so if k constant from Z is k then for D length it is k(m1+m2)
so period is 2 p square root from m2/k(m1+m2)
is that right

No. Check the parentheses and dimensions. I do not understand the sentence "if k constant from Z is k then for D length it is k(m1+m2)"

 

[giokrutoi]

No. Check the parentheses and dimensions. I do not understand the sentence "if k constant from Z is k then for D length it
is k(m1+m2)"

sorry i made mistakes there
here in attached file I wrote correct

 

[ehild]

sorry i made mistakes there
here in attached file I wrote correct

I can not read your attachment. Type in your work.

 

[giokrutoi]

I can not read your attachment. Type in your work.

if spring has length Z and i call D length from m2 to center of mass then m1(Z-D)=m2D
D is m1Z/(m1+m2) now spring constant is untiproportional to length so k1 (new constant) Times D equal k times Z so K1=(m1+m2)k/m1
so period is 2 pi square root from m1m2/k(m1+m2)

EDIT : Is that right

 

[ehild]

The spring constant of which piece is k1?
The period is correct.

 

[giokrutoi]

The spring constant of which piece is k1?
The period is correct.

from center of mass to m2

 

[ehild]

OK, but would not it be more logical to call it k2? And what is the spring constant of the other piece and the period of mass m1?

 

[giokrutoi]

OK, but would not it be more logical to call it k2? And what is the spring constant of the other piece and the period of mass m1?

i calculated and it's same

EDIT :Now acceleration is omega squared times maximal strechment so omega squared is m1m2/(m1+m2)k
all I have to do is to calculate maximal strechment is that right?

 

[ehild]

EDIT :Now acceleration is omega squared times maximal strechment so omega squared is m1m2/(m1+m2)k
all I have to do is to calculate maximal strechment is that right?

Yes.

 

[giokrutoi]

hey I think I figured out everything
look
we can calculate maximal speed of m1 at the release point from conservation of energy all the potential energy is converted in kinetic of m1
then this v is maximal for m1 thing is that m1's coordinate dependence on time look's so
y=XmaxSINUS OMEGA t were we know omega
t is time Xmax is maximal compression
now if we take derivative from that equation we will get velocity dependence on time
v=VmaxCOSINUS omega t
vmax is maximal velocity of m1 in reference frame of center of mass so we have to subtract velocity of center of mass from v maximal which we defined by conservation of energy
vmax also equals XMAX times omega so we know vmax and omega and we can find Xmax

now acceleration of first mass is omwga squared vmax we know both omega and vmax so we can find acceleration a1

then m1a1=m2a2 a2=m1a1/m2
is that conclusion right

 

[giokrutoi]

?

 

[haruspex]

hey I think I figured out everything
look
we can calculate maximal speed of m1 at the release point from conservation of energy all the potential energy is converted in kinetic of m1
then this v is maximal for m1 thing is that m1's coordinate dependence on time look's so
y=XmaxSINUS OMEGA t were we know omega
t is time Xmax is maximal compression
now if we take derivative from that equation we will get velocity dependence on time
v=VmaxCOSINUS omega t
vmax is maximal velocity of m1 in reference frame of center of mass so we have to subtract velocity of center of mass from v maximal which we defined by conservation of energy
vmax also equals XMAX times omega so we know vmax and omega and we can find Xmax

now acceleration of first mass is omwga squared vmax we know both omega and vmax so we can find acceleration a1

then m1a1=m2a2 a2=m1a1/m2
is that conclusion right

Possibly, it's hard to be sure without seeing your working and answer. Keeping all the given data symbolic, (m₁ etc.) what expression do you get for the max acceleration of m₂?
By the way, you are only given numeric values for initial compression and the two masses, not the spring constant, right? That leaves you without enough information to get a numeric answer for acceleration.

 

[giokrutoi]

not the spring constant, right?

no I have spring constant 20n/m.i forgot to mention it in early posts I said that k is given

 

[ehild]

hey I think I figured out everything
look
we can calculate maximal speed of m1 at the release point from conservation of energy all the potential energy is converted in kinetic of m1
then this v is maximal for m1 thing is that m1's coordinate dependence on time look's so
y=XmaxSINUS OMEGA t were we know omega

That Vmax is valid in the ground frame of reference. After m2 is released, the whole system translates with the velocity of the center of mass, so both objects move with that velocity + the velocity of their vibration.
You turned to the center of mass frame of reference, where the blocks oscillate , but the maximum speed of oscillation is different from Vmax you got previously. You have to subtract the velocity of the center of mass.
If you got it, you can use that the maximum acceleration is maximum speed multiplied by omega. that you know already.

 

[giokrutoi]

That Vmax is valid in the ground frame of reference. After m2 is released, the whole system translates with the velocity of the center of mass, so both objects move with that velocity + the velocity of their vibration.
You turned to the center of mass frame of reference, where the blocks oscillate , but the maximum speed of oscillation is different from Vmax you got previously. You have to subtract the velocity of the center of mass.
If you got it, you can use that the maximum acceleration is maximum speed multiplied by omega. that you know already.

yes that's exactly what I was saying
thanks a lot. For waisting your time on. Helping me

 

[haruspex]

yes that's exactly what I was saying
thanks a lot. For waisting your time on. Helping me

Just to check, the max acceleration I got for m₂ is $\frac{kx_1}{\sqrt{m_2(m_1+m_2)}}$

 

[ehild]

Just to check, the max acceleration I got for m₂ is $\frac{kx_1}{\sqrt{m_2(m_1+m_2)}}$

It depends what x₁ means. It should be the initial compression, 0.1 m.

 

[ehild]

It depends what x₁ means. It should be the initial compression, 0.1 m.

I outline the energy method suggested by jbriggs and me :
ΔL = 0.1 m is the initial compression of the spring.
V1 is the maximum speed of block1 after it was released, but block2 still fixed.
The total energy of the system is 1/2 m₁V₁²=1/2 k (ΔL)²
$V_1 = \sqrt \frac {k}{m_1}\Delta L$
Speed of the CoM $V_{CoM}= \frac {m_1V_1}{m_1+m_2}$
energy of translation $E_{tr}=1/2 (m_1+m_2 ) \left( \frac {m_1V_1}{m_1+m_2} \right)^2$.
Let be S the maximum distortion of the spring after both blocks were released.
The spring energy is $E_s =E-E_{tr}= 1/2 V_1^2 \frac {m_1m_2}{m_1+m_2}= 1/2 k S^2$
$S=V_1\sqrt {\frac {m_1m_2}{k(m_1+m_2)}}$ and a₂=KS/m₂.

 

[haruspex]

I outline the energy method suggested by jbriggs and me :
ΔL = 0.1 m is the initial compression of the spring.
V1 is the maximum speed of block1 after it was released, but block2 still fixed.
The total energy of the system is 1/2 m₁V₁²=1/2 k (ΔL)²
$V_1 = \sqrt \frac {k}{m_1}\Delta L$
Speed of the CoM $V_{CoM}= \frac {m_1V_1}{m_1+m_2}$
energy of translation $E_{tr}=1/2 (m_1+m_2 ) \left( \frac {m_1V_1}{m_1+m_2} \right)^2$.
Let be S the maximum distortion of the spring after both blocks were released.
The spring energy is $E_s =E-E_{tr}= 1/2 V_1^2 \frac {m_1m_2}{m_1+m_2}= 1/2 k S^2$
$S=V_1^2\sqrt {\frac {m_1m_2}{k(m_1+m_2)}}$ and a₂=KS/m₂.

Yes, that looks to be what I did too. No need to get into the SHM details.

 

[giokrutoi]

thanks very much