Spring-mass problem

Question

Question: A spring-mass system hangs vertically from a fixed support. The natural length of the spring is l. The block is displaced from the equilibrium position and constrained to move horizontally. If the displacement is small, show that the potential energy of the system is proportional to the fourth power of the displacement. Also find the relation between the time period and the displacement.

My attempt: Spring force $F=-k(\sqrt{l^2 + x^2} -l)$

Horizontal component of spring force = $ \frac{Fx}{\sqrt{l^2 + x^2}} $

Integrating this should give the negative of the potential energy function U.

Thus, on integration, $U=\frac{x^2}{2} - l\sqrt{l^2 + x^2}$

However, this is not proportional to $x^4$.

So where is the fallacy?

Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. — John Rennie, Feb 22 '18 at 15:46
If the displacement is small ... You have not applied this approximation. — sammy gerbil, Feb 22 '18 at 17:45

score 0 · Answer 1 · answered Feb 22 '18 at 16:43

For Forces which are linear function of displacement such as springs i reckon it is impossible to have potential energy proportional to 4th power of displacement. This i concluded from the property of conservative force itself. The negative derivative of potential energy must be equal to force, if PE depends on 4th power of displacement then force should be of 3rd power which never occurs in case of springs. -Correct me for any mistakes.

Spring-mass problem

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