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While attempting problems of simple harmonic motion I came across this problem, which has gotten me confused.

A fixed horizontal spring is stretched by a constant force $F$. I am required to obtain the maximum elongation of that spring. But the problem is which method is correct, the energy method or the force method? Do let me know the misconception in the wrong path.

Method 1:

$$\text{For Equilibrium}\\ \; F= kx_\text{max}\\ \implies x_\text{max}= \frac{F}{k}\;.$$

Method 2:

$$\text{Work done by the force}\; = \text{change in potential energy of the spring}\\ \text{i.e.} \; F\cdot x_\text{max}= \frac{1}{2} kx_\text{max}^2\\ \implies x_\text{max} = \frac{2F}{k}\; .$$

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    $\begingroup$ Even in ideal cases, there is a chance of losing energy as heat. The same happens in case of RC circuits, half the energy is lost as heat & the other half is stored in the capacitor. $\endgroup$
    – Yashas
    Nov 16 '15 at 17:18
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    $\begingroup$ @YashasSamaga your comment is unrelated to the question and might confuse the OP $\endgroup$
    – user83548
    Nov 16 '15 at 17:40
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    $\begingroup$ Basically, which answer is the correct one depends on how you interpret the question. If it is held a rest by the constant force, the force method is correct, if you have a constant force along the path and the system oscillates, the energy method is the correct one $\endgroup$
    – user83548
    Nov 16 '15 at 17:43
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    $\begingroup$ Your main problem was to consider $F$ as constant. Restoring force is increasing as you displace the spring away from the equilibrium. $\endgroup$
    – user36790
    Nov 16 '15 at 17:51
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    $\begingroup$ The statement of the question is ambiguous. Is it being held by a constant force, or is it being actively stretched by a constant force. $\endgroup$
    – garyp
    Nov 16 '15 at 19:08
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The first method is giving the correct answer. In writing the work done by the force, you are assuming that the force $F$ itself is constant throughout the extension. However, this is not true. While extending the spring in a quasi-static way, the force $F$ must always match exactly the spring force at that time. This is needed so that at the end of the extension, the spring remains at rest. Once we understand this, we note that the force at extension $x$ is $F(x) = k x$. Then, the work done along the path is $$ \int_0^{x_{\max}} F(x) dx = \frac{1}{2} k x_\max^2 $$ The latter of course is precisely the potential energy of the spring. Thus, the "energy conservation" equation is trivial and does not yield any new information.

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  • $\begingroup$ But it is stated that a constant force is applied. $\endgroup$
    – Abhinav
    Nov 16 '15 at 17:20
  • $\begingroup$ if you assume a constant force the spring would still reach a maximum extent at zero speed, what is it wrong with that? $\endgroup$
    – user83548
    Nov 16 '15 at 17:21
  • $\begingroup$ What you say does make sense.I suppose the author might have missed out on that $\endgroup$
    – Abhinav
    Nov 16 '15 at 17:21
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    $\begingroup$ It is stated that a constant force is used to hold the spring in place. It is not stated whether a constant force was used to bring the spring into its current state. You made the latter assumption. $\endgroup$
    – Prahar
    Nov 16 '15 at 17:21
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    $\begingroup$ @Prahar I read the question again and I agree with you it can be interpreted both ways $\endgroup$
    – user83548
    Nov 16 '15 at 17:36
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If you use a constant force along the path, the spring will move past the position where $F=kx$, because it will reach that point at some speed. Thus it is incorrect to use the force method in the way you used it, because at maximal extension $v=0$ but $a\neq0$. The energy method as you used it will give the correct answer. If, instead, the force is used to keep the spring elongated at rest, then the force method is correct. To see why read Prahar's answer.

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  • $\begingroup$ Yes, this is correct. +1. $\endgroup$
    – user36790
    Nov 16 '15 at 18:07

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