Springs of spring constants k, 2k, 4k, 8k ... 2048 k are connected in series. A mass ‘m’ is attached to one end the system is allowed to oscillate. Its time period is approximately

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2 π \sqrt{\frac{m}{2K}}

2 π \sqrt{\frac{4m}{K}}

2 π \sqrt{\frac{m}{4K}}

2 π \sqrt{\frac{2m}{K}}

answer is B.

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Detailed Solution

\frac{1}{k_{eff}} = \frac{1}{k} + \frac{1}{2k} + \frac{1}{4k} + \frac{1}{8k} + - - + \frac{1}{2048 k}

$= \frac{1}{k} [1 + \frac{1}{2} + \frac{1}{2^{2}} + - - + \frac{1}{2^{11}}]$

$S_{n} = a_{1} [\frac{1 - r^{n}}{1 - r}]$

Where $a_{1}$ is 1^st term and n = 4

$r = \frac{1}{2}$

$\frac{1}{k_{eff}} = \frac{1}{k} (\frac{(1 - {(\frac{1}{2})}^{11})}{(1 - \frac{1}{2})})$

$\frac{1}{k_{eff}} = \frac{1}{k} (\frac{2^{11}}{2^{10}}) = \frac{2}{k}$

$T = 2 π \sqrt{\frac{m}{k_{eff}}} = 2 π \sqrt{\frac{2m}{k}}$

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