A simple spring has length $l$ and force constant k. It is cut into two spring of length $l_{1}andl_{2}\ such\ that\ l_{1} = nl_{2}(\ n = an\ integer).$ the force constant of the spring of length $l_{2}\ is\ $

Question

Dr. Abubacker · Accepted Answer

(b) Let $k$ be the force constant of spring of length$l_{2}$. Since $l_{1} = n\ l_{2}$ , where $n$ is an integer, so the spring is made of $(n + 1)$ equal parts in length each of length $l_{2}$
$	herefore\ \frac{1}{k} = \frac{(n + 1)}{k}\ \ \ or\ \ \ k = (n + 1)k$
The spring of length $l_{1}( = n\ l_{2})$ will be equivalent to $n$ springs connected in series where spring constant $k^{'} = \frac{k}{n} = \frac{(n + 1)k}{n}$

Q. A simple spring has length $l$ and force constant k. It is cut into two spring of length $l_{1}andl_{2}\ such\ that\ l_{1} = nl_{2}(\ n = an\ integer).$ the force constant of the spring of length $l_{2}\ is\ $

Explanation

Reviewed by Dr. Abubacker

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