Hooke's law spring configurations

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Homework Help Overview

The problem involves analyzing the behavior of springs under different mass configurations, specifically using Hooke's law. It compares a single mass hanging from a spring to two smaller masses hanging from a spring, both with the same spring constant.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of treating the center of the spring as fixed and explore the concept of symmetry in the setup. Some suggest simplifying the problem by focusing on one half of the spring or considering the spring fixed at one end. Others express confusion about the necessity of these approaches and question the assumptions involved.

Discussion Status

The discussion is active with various perspectives being explored. Some participants have offered guidance on simplifying the problem, while others are questioning the assumptions made in the original post. There is no explicit consensus, but multiple interpretations and approaches are being considered.

Contextual Notes

Participants note that the problem is part of an online quiz, which may impose specific constraints on how answers are formatted or presented. There is also mention of the challenge of applying Hooke's law after a period of time away from the subject.

judas_priest
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Homework Statement


As shown in Figure A, a block of mass m is hanging from a spring attached to the ceiling. As shown in Figure B, two blocks of mass m/2 are hanging from two strings that are attached to a spring that has the same spring constant k. If the spring in Figure A is stretched a distance d, how far will the spring in Figure B stretch?2. The attempt at a solution

The problem (Figure-B)is approached based on identifying that the center of the spring acts like a fixed point and does not shift (because of symmetry). Therefore the problem now breaks down into two equal springs in series. Don't know how to approach from here. After it breaks down into two springs of equal spring constants, I used [itex](k1k2)x/2(k1+k2) = mg/2[/itex] did this for both masses respectively. Not getting me anywhere. Anyone have a different approach or another way to look at this problem?
 

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HINT: The elongation depends on tension in the spring.
 
judas_priest said:
the problem now breaks down into two equal springs in series. Don't know how to approach from here. After it breaks down into two springs of equal spring constants, I used [itex](k1k2)x/2(k1+k2) = mg/2[/itex] did this for both masses respectively.
That's not quite the point of taking the centre to be fixed. The idea is that you only have to think about one half of the spring and one mass. You can conceptually replace the other half of the system by a wall.
 
haruspex said:
That's not quite the point of taking the centre to be fixed. The idea is that you only have to think about one half of the spring and one mass. You can conceptually replace the other half of the system by a wall.

Is that even necessary? I mean the problem can be without cutting the spring.
 
Make the problem simple. If the system is static, consider the spring fixed or clamped at one of the pulleys. Then compare the tension in the spring in case A with case B. What causes the tension, and how much is it? No formulas needed.
 
haruspex said:
That's not quite the point of taking the centre to be fixed. The idea is that you only have to think about one half of the spring and one mass. You can conceptually replace the other half of the system by a wall.

But that would just give me [tex]T = Mg/2[/tex] and substituting for M from first system would give me [tex]T = kd/2[/tex]. Substituting T as [tex]-kx[/tex] would give me [tex]x = d/2[/tex]

Is this correct?
 
barryj said:
Make the problem simple. If the system is static, consider the spring fixed or clamped at one of the pulleys. Then compare the tension in the spring in case A with case B. What causes the tension, and how much is it? No formulas needed.

Wouldn't the tension on both sides be same? Because the string is ideal, and hence tension is the same everywhere. Or did I not get your point right?
 
I can't believe you guys are making the problem so complicated. If you fix the spring at the right end, then have the same spring but with half the mass.
 
darkxponent said:
Is that even necessary? I mean the problem can be without cutting the spring.

So, what's your way of looking at it and approach to it?
 
Last edited:
  • #10
barryj said:
I can't believe you guys are making the problem so complicated. If you fix the spring at the right end, then have the same spring but with half the mass.

Yup, it's quite a simple problem. Just getting back to Hooke's law after 2 years.
 
  • #11
barryj said:
I can't believe you guys are making the problem so complicated. If you fix the spring at the right end, then have the same spring but with half the mass.
My reading of the OP is that symmetry is the ordained method to be used.
 
  • #12
judas_priest said:
So, what's your way of looking at it and approach to it?

For me the problem is even simpler. Just find the Tension in the spring and equating it to Kx. There is no need to fix the strung or cut the spring!
 
  • #13
darkxponent said:
For me the problem is even simpler. Just find the Tension in the spring and equating it to Kx. There is no need to fix the strung or cut the spring!

That was my initial attempt. Doesn't get me to an answer.
Anyone have a different approach?
 
  • #14
haruspex said:
My reading of the OP is that symmetry is the ordained method to be used.

Could you elaborate?
 
  • #15
Oh, man. This was a simple problem. Turns out I was entering the answer right with the wrong syntax. It's an online quiz I'm taking. D/2 is the answer. I was putting parentheses around, and hence getting it wrong.
 

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