# JEE Main & Advanced Physics Simple Harmonic Motion Spring System

Spring System

Category : JEE Main & Advanced

When a spring is stretched or compressed from its normal position $(x=0)$ by a small distance $x,$ then a restoring force is produced in the spring because it obeys Hook's law

i.e.   $F\propto -x$$\Rightarrow$$F=-\,k\,x$

where k is called spring constant.

(i) It's S.I. unit Newton/metre, C.G.S unit Dyne/cm and dimension is $[M{{T}^{-2}}]$

(ii) Actually k is a measure of the stiffness/softness of the spring.

(iii) For massless spring constant restoring elastic force is same every where

(iv) When a spring compressed or stretched then work done is stored in the form of elastic potential energy in it.

(v) Spring constant depend upon radius and length of the wire used in spring.

(vi) The spring constant k is inversely proportional to the spring length.

$k\propto \frac{1}{\text{Extension}}\propto \frac{1}{\text{Length}\,\text{of spring}}$

That means if the length of spring is halved then its force constant becomes double.

(vii) When a spring of length $l$ is cut in two pieces of length ${{l}_{1}}$ and ${{l}_{2}}$ such that ${{l}_{1}}=n{{l}_{2}}$.

If the constant of a spring is k then spring constant of first part

${{k}_{1}}=\frac{k(n+1)}{n}$

Spring constant of second part  ${{k}_{2}}=(n+1)k$

and ratio of spring constant $\frac{{{k}_{1}}}{{{k}_{2}}}=\frac{1}{n}$

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