What Determines the Maximal Elongation of a Spring in a Pulley System?

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SUMMARY

The discussion focuses on the mechanics of a pulley system involving two bricks of mass m connected by an ideal weightless spring with spring constant k. The key conclusion is that the minimal time τ for maximal elongation ΔL of the spring occurs when the amplitude A0 is determined to be mg/3k. This is derived from analyzing the system's acceleration and forces acting on the blocks, particularly in a non-inertial frame where the average acceleration is g/3. The solution emphasizes the importance of understanding the dynamics of the system to accurately calculate elongation.

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snate
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Homework Statement


2 bricks of mass m are located near the edge of a smooth horizontal surface, they are connected by an ideal weightless ,unstretched spring with a length of l0 and spring constant k. The brick which is closer to the edge is connected to another brick of the same mass by an ideal unstretchable rope going through a pulley. The lowest brick is held so that the rope above it hangs vertically.
The lowest brick is released
What's the minimal time τ after which the elongation of the spring ΔL will be maximal. Find ΔL.

20645861_1928917583989680_1368339303_n.png
20631554_1928917840656321_1206854442_n.png

Homework Equations

The Attempt at a Solution


Here's the solution until the point where I'm stuck. Sorry for the Russian text, it does not contain crucial information, please ignore it
20632389_1928930010655104_1002110814_n.png

a2-a1 is the acceleration of one end of the spring from the frame of reference of another end.
Then these equations are given (y is vertical coordinate)
20631623_1928927060655399_1063382409_n.png

I understand why w2 is 3k/2m but why is the amplitude(A0) mg/3k? Can someone please explain? I only need an explanation for A0
The next equations are
20631814_1928927260655379_432063850_n.png

20632625_1928928830655222_42064552_n.png

20631703_1928928893988549_190674348_n.png

At the beginning the system is in equilibrium and then x(0)=0, that leads to B being equal to 0 and and A0+A=0 because the spring isn't stretched at the beginning.
20632455_1928929140655191_236803601_n.png

20643868_1928929220655183_226754705_n.png

20632337_1928929310655174_1726256519_n.png

Thanks.
 
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I have not tried to follow your equations, but I can suggest an easier way via a sequence of simplifications.
First, model it as all happening on a horizontal surface, with a horizontal force mg applied to the rightmost block.
Next, we can merge the two blocks connected by the inextensible string into a single block mass 2m.
Clearly the average acceleration is g/3 to the right, so take a noninertial frame accelerating that way. In this frame, the left-hand block has a pseudoforce ... to the left, and the 2m block has a pseudoforce ... to the left, leaving it with a net force ... to the right. (Fill in the blanks.)
Finally, consider the common mass centre of the two blocks. How does this move in the reference frame? How do the two blocks move in relation to it?
 
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haruspex said:
I have not tried to follow your equations, but I can suggest an easier way via a sequence of simplifications.
First, model it as all happening on a horizontal surface, with a horizontal force mg applied to the rightmost block.
Next, we can merge the two blocks connected by the inextensible string into a single block mass 2m.
Clearly the average acceleration is g/3 to the right, so take a noninertial frame accelerating that way. In this frame, the left-hand block has a pseudoforce ... to the left, and the 2m block has a pseudoforce ... to the left, leaving it with a net force ... to the right. (Fill in the blanks.)
Finally, consider the common mass centre of the two blocks. How does this move in the reference frame? How do the two blocks move in relation to it?
Thanks, your explanation hinted me why A0 is mg/3k. A0=amax/w2. And maximal acceleration amax after infinitesimal moment of time since the rightmost block has been released is gm/2m=g/2 because at that moment the spring is not deformed, so F=kΔx=k0=0, so we don't have to consider the spring and the leftmost block at that moment. So A0 =g*2m/(2*3*k)=mg/3k. Am I correct?
 
snate said:
Thanks, your explanation hinted me why A0 is mg/3k. A0=amax/w2. And maximal acceleration amax after infinitesimal moment of time since the rightmost block has been released is gm/2m=g/2 because at that moment the spring is not deformed, so F=kΔx=k0=0, so we don't have to consider the spring and the leftmost block at that moment. So A0 =g*2m/(2*3*k)=mg/3k. Am I correct?
Looks good.
 
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