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