Rankers Physics

Spring Block System: Practice Problem & Solution

A spring of force constant $k$ is cut into lengths of ratio $1 : 2 : 3$. They are connected in series and the new force constant is $K'$. Then they are connected in parallel and force constant is $K''$. Then $K' : K''$ is: (2017-Delhi)
$1 : 9$
$1 : 11$
$1 : 14$
$1 : 6$

Solution Explained:

To solve this problem, we apply the core principles of Spring Block System. Understanding the underlying formula is key to arriving at the correct answer below:

$k \propto 1/L$. The parts have stiffness $6k, 3k, 2k$. In series, $K' = k$. In parallel, $K'' = 6k + 3k + 2k = 11k$. The ratio is $1:11$.

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