# Spring-mass problem [duplicate]

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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?

## marked as duplicate by John Rennie, sammy gerbil, Emilio Pisanty, Jon Custer, Chris♦Feb 24 '18 at 0:12

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## 1 Answer

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.