Question

A spring of force constant $K$ is cut into two pieces such that one piece is double the length of the- other, then find out the force constant of the longer piece $(K=6N/m)$ in $N/m$.

Answer 1

Let the length of the spring be L and spring constant be k

Now, the larger piece will be$\frac{2L}{3}$

We know that,

$k$ is inversely proportional to length of the spring

Now,

$k_1=\frac{a}{L}...\left(I\right)$

$k_2=\frac{a}{\frac{2L}{3}}....\left(II\right)$

Now, dividing both equation

$\frac{k_1}{k_2}=\frac{a\times 2L}{L\times 3a}$

$\frac{k_1}{k_2}=\frac{2}{3}$

$k_2=\frac{3k_1}{2}$

$k_2=\frac{3\times 6}{2}$

$k_2=9N/m$

Hence, this is the required solution