Infinite springs with force constants k,2k,4k | vedclass

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Question

(C) For springs connected in series,the effective force constant $k_{eff}$ is given by the formula: $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}

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$\frac{k}{2}$

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(C) For springs connected in series,the effective force constant $k_{eff}$ is given by the formula: $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} + \dots$Substituting the given values: $\frac{1}{k_{eff}} = \frac{1}{k} + \frac{1}{2k} + \frac{1}{4k} + \frac{1}{8k} + \dots$Factor out $\frac{1}{k}$: $\frac{1}{k_{eff}} = \frac{1}{k} \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots ight)$The term in the bracket is an infinite geometric progression with first term $a = 1$ and common ratio $r = \frac{1}{2}$.The sum $S$ of an infinite geometric progression is given by $S = \frac{a}{1 - r}$.$S = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.Therefore,$\frac{1}{k_{eff}} = \frac{1}{k} 	imes 2 = \frac{2}{k}$.Thus,$k_{eff} = \frac{k}{2}$.

Infinite springs with force constants $k$,$2k$,$4k$,$8k$,... respectively are connected in series. The effective force constant of the system will be:

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