How Does Adding More Springs Affect Extension in Hooke's Law Experiments?

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Adding more springs affects extension in Hooke's Law experiments by altering the effective spring constant. When springs are connected in series, the overall spring constant decreases, leading to increased extension for the same force. Conversely, when springs are arranged in parallel, the effective spring constant increases, resulting in reduced extension. The discussion highlights that the relationship between force, spring constant, and extension is crucial for understanding these configurations. Overall, the experiment emphasizes the importance of proper setup and understanding the underlying physics principles.
thang08
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Homework Statement



This isn't acctually a problem but I really don't know what to do. I was doing a simple activity about Hooke's Law during class today where I have to set up a apparatus and measure the extension of spring when when we put differnt masses on it. I didn't do it properly so I have difficulty answering it

This is my graph if it helps any

http://img156.imageshack.us/img156/8913/clipuw9.jpg
you see, I only did the experiment with 1 spring. I didn't do the following required experiments, so I have no idea how to answer.

Homework Equations



Q6. Write how much extension do you think there would be (compared to the extension for one spring) if you now had two springs, identical to the one used, joined one after the other

Q7. Observe what happens (describe). Explain

Q8. Do the same, but this time for two springs side by side

Q9 What about three side by side, or four, or three in a row, or four?

The Attempt at a Solution



Q6. I think it remains the same because F and k remains the same => x remains
Q7. Observation would suit what I just said
Q8. 2 springs side by side so forced is distributed even onto both springs thus making the extension halved
Q9. I don't even understand the question

Thanks in advance:!)
 
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I like q8 response a lot!

So let's call a new spring constant K'eq that which satisfies the eqn,

f=-kx. the force is the same, and as you propose extention is halved,
f=-keq*.5x divide be
f=-kx and you will find k'eq=2k. For q9 extend your logic for 3 or more springs.

now q6 is a bit more subtle, but if you think more about it, you may start to doubt your answer.
 
I don't get why Q6 response is wrong, well as far as I understand force constant of a spring shows us how stiff it is. If two identical springs connected in 'series', the stiffness of the 'combined' spring doesn't change. If we don't increase the mass, i.e. no change in weight force, the extension (x) of the spring shouldn't change.
 
err, no suggestion? T_T I keep on think my way of thinking is right and some online experiment tells me that it's wrong. Explanation please
 
Been gone, think about this thought experiment forget on-line proof etc. You have a bungee like cord and want to stretch it by 2 inches. Cut the bungee cord in half and see how much more effort it requires to stretch the same 2 inches. Or if you want to save yourself a buck and have some rubberbands handy, for grins tie 4 in series and compare that with one in terms of stretch force and displacement.
The math is relatively trivial thing to prove, but I'm thinking it might be more helpful to get it out of the purely math realm. Hope this helps a bit.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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