A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of

Question

A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of (A) (2 k/3) (B) (3 k/2) (C) 3k (D) 6 k

Tardigrade Admin · Accepted Answer

l1=2l2 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-dpdxwoblk6im8jvp.jpg" /> Therefore l 1 = (2/3) l â¡ Force constant, k propto (1/ textlength of spring) ⇒ k1=(3/2)k

Q. A spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of

$\frac{2 k}{3}$

$\frac{3 k}{2}$

$3 k$

$6 k$

$l_{1} = 2 l_{2}$

Therefore
$l_{1} = \frac{2}{3} l$
Force constant,
$k \propto \frac{1}{length of spring}$

$\Rightarrow k_{1} = \frac{3}{2} k$

Q. A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of

32k​

23k​

3k

6k