A spring has length L and force constant K . It is cut into two springs of length L 1 and L 2 such that L 1 = NL 2 ( N is an integer). The force constant of spring of length L 1 is

Question

A spring has length L and force constant K . It is cut into two springs of length L 1 ​ and L 2 ​ such that L 1 ​ = NL 2 ​ ( N is an integer). The force constant of spring of length L 1 ​ is

TestHub · Accepted Answer

The force constant of a spring is inversely proportional to its length, so $K \propto \frac{1}{L}$ .

Let the original length of the spring be $L = L_{1} + L_{2}$ .

Given $L_{1} = N L_{2}$ , we have $L_{2} = \frac{L _{1}}{N}$ .

Substituting this into the equation for $L$ , we get $L = L_{1} + \frac{L _{1}}{N} = L_{1} (\frac{N + 1}{N})$ .

Since $K_{1} L_{1} = K L$ , the force constant of the spring of length $L_{1}$ is $K_{1} = K \frac{L}{L _{1}}$ .

Substituting the value of $L$ , we get $K_{1} = K \frac{L _{1} ( \frac{N + 1}{N} )}{L _{1}} = \frac{K}{N} (1 + N)$ .

Answer: $\frac{K}{N} (1 + N)$

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