Question

A block of weight $W$ produces an extension of $9\, cm$ when it is hung by an elastic spring of length $60\, cm$ and is in equilibrium. The spring is cut into two parts, one of length $40\, cm$ and the other of length $20\, cm$. The same load $W$ hangs in equilibrium supported by both parts as shown in the figure. Find the extension (in $cm$ ) in the spring.

Answer 1

Let $k$ be the spring constant of the original spring
$W=9 k \Rightarrow k=\frac{W}{9}$
Spring constant is inversely proportional to the length of the spring.
Spring constant for the $40 cm$ piece
$k_{1}=k \times \frac{60}{40}=\frac{3}{2} k$,
$k_{2}=k \times \frac{60}{20}=3\, k$
When connected in parallel, the equivalent spring constant
$k'=k_{1}+k_{2} $
$\Rightarrow k'=\frac{3}{2} k+3 k=4 \frac{1}{2} k$
Let $x$ ' be the new extension.
Then $k' x'=W=9\, k$
$\Rightarrow \frac{9}{2} k \times x'=9\, k$
$\Rightarrow x'=9 k \times \frac{2}{9 k}=2\, cm$