Answer:
For a spring \[\text{k}\propto
\frac{1}{l}\]. \[\therefore \]\[{{\text{k}}_{\text{1}}}\propto
\frac{1}{{{l}_{1}}}\], \[{{\text{k}}_{\text{2}}}\propto \frac{1}{{{l}_{2}}}\] and \[\text{k}\propto
\frac{1}{{{l}_{1}}+{{l}_{2}}}\]
\[\therefore
\]\[\frac{{{k}_{1}}}{k}=\frac{{{l}_{1}}+{{l}_{2}}}{{{l}_{1}}}=1+\frac{{{l}_{2}}}{{{l}_{2}}}\] or \[{{\text{k}}_{\text{1}}}=\text{
k}\left( 1+\frac{{{l}_{2}}}{{{l}_{1}}} \right)\]
and \[\frac{{{k}_{2}}}{k}=\frac{{{l}_{1}}+{{l}_{2}}}{{{l}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}+1\] or \[{{\text{k}}_{2}}=\text{
k}\left( 1+\frac{{{l}_{1}}}{{{l}_{2}}} \right)\]
Note : If \[{{\text{l}}_{\text{1}}}={{\text{l}}_{\text{2}}},\text{
}{{\text{k}}_{\text{1}}}=\text{ }{{\text{k}}_{\text{2}}}=\text{2}k\] i.e. when
length is halved, spring constant of each half becomes twice.
If \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{1}{n}\], then \[{{\text{k}}_{\text{1}}}=\text{
k}\left( \text{l }+\text{ n} \right);{{\text{k}}_{\text{2}}}=\text{ k}\left(
1+\frac{1}{n} \right)\]
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