0
$\begingroup$

enter image description here

Consider a vertical spring on which we hang a mass m. It will stretch a distance Δx because of the weight of the mass. Suppose, at this position the mass is at equilibrium (mg=kΔx).

Now if I pull the mass down an additional distance x', the mass will start oscillating and its motion will be 'Simple Harmonic Motion.

In my textbook, it is written that in this case, the additional stretch, x' must be less than the spring's initial stretch due to the weight of the mass, Δx.

Why x' has to be less than Δx in this case? What would be the problem if x' was equal or greater than Δx?

|cite|improve this question
$\endgroup$
  • $\begingroup$ There's no problem ... unless the unstretched spring is of the kind where the coils are completely compressed and in contact with their neighbors. Springs like that are common in teaching laboratories. If $x'>\Delta x$ the spring would "hit the top" at the top of it's travel. That's all I can think of. $\endgroup$ – garyp Jul 10 '17 at 11:51
  • $\begingroup$ What will happen if the spring is of the kind you mentioned? $\endgroup$ – Hisab Jul 10 '17 at 12:11
  • $\begingroup$ @JMac answer is more complete than mine, which is an extreme case of an extension spring. In my picture the compressive spring constant is infinite. The textbook (or lab manual if it is one) should have made some comment about this. If it were a textbook, and it did not make a comment, you should throw the book in the trash. $\endgroup$ – garyp Jul 11 '17 at 11:19
  • $\begingroup$ Please, suggest me some good textbooks. $\endgroup$ – Hisab Jul 11 '17 at 17:09
1
$\begingroup$

The problem will be in the motion of the spring.

In simple harmonic motion the spring will go between it's maximum and minimum height. For it to be "simple" we ignore air resistance/any losses and assume the spring is massless and perfectly Hookean ($F = k \Delta x$).

For this simple harmonic motion; the mass will oscillate around the equilibrium position (stretched length when not moving). It will go the same distance above the equilibrium position as it will go below the equilibrium position.

An important thing to note is that for an extension spring; it will not behave properly if you compress it below it's unstretched length (it is designed to only act as a perfect spring in extension).

If you displace it further than the mass displaces it, it will try and compress the spring smaller than it's unstretched length, where it would no longer behave as a perfect spring and simple harmonic motion would not work.

If the spring could operate in extension and compression and had the same k in both directions, it would not be a problem as long as minimum length was lower than the minimum length you would compress it to.

Basically, this is only true if this was an extension spring that could not go shorter than it's unstretched length.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks but couldn't understand this paragraph : //If the spring could operate in extension and compression and had the same k in both directions, it would not be a problem as long as minimum length was lower than the minimum length you would compress it to.// $\endgroup$ – Hisab Jul 10 '17 at 13:40
  • $\begingroup$ @Hisab Basically I'm assuming that the spring isn't designed to compress smaller than it's unstretched length. It is possible to design a spring that can do that, and in that case as long as $k$ was the same for that compression, you would be able to displace it more than that. $\endgroup$ – JMac Jul 10 '17 at 13:42
  • $\begingroup$ Okay, but I have problem with this line: //it would not be a problem as long as minimum length was lower than the minimum length you would compress it to.// If the spring can compress smaller than it's unstretched length, no problem should remain. But you are applying another condition which I don't understand. Will you please, explain? $\endgroup$ – Hisab Jul 10 '17 at 13:49
  • $\begingroup$ @Hisab Yes, exactly. Your textbook seems to be talking about a specific case where we assume the spring cannot compress below it's unstretched length. The problem only exists under certain assumptions that the book doesn't make clear. $\endgroup$ – JMac Jul 10 '17 at 13:50
  • $\begingroup$ I have already understood that, thank you. I don't understand that one line: //it would not be a problem as long as minimum length was lower than the minimum length you would compress it to.// What do you mean by minimum length here? $\endgroup$ – Hisab Jul 10 '17 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.