If two similar springs each of spring constant K1 are joined in series, the new spring constant and time period would be changed by a factor :
Solution
${1 \over {{K_{eq}}}} = {1 \over {{K_1}}} + {1 \over {{K_1}}}$<br><br>$$ \Rightarrow {K_{eq}} = {{{K_1} \times {K_1}} \over {{K_1} + {K_2}}} = {{K_1^2} \over {2{K_1}}} = {{{K_2}} \over 2}$$<br><br>$\therefore$ $T' = 2\pi {\sqrt {{m \over {{K_{eq}}}}} }$<br><br>$= 2\pi {\sqrt {{m \over {{{{K_1}} \over 2}}}} }$<br><br>$= 2\pi {\sqrt {{{2m} \over {{K_1}}}} }$<br><br>$= \sqrt 2 T$ [ where $T = 2\pi {\sqrt {{m \over {{K_1}}}} }$]
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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