JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 10)
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 :
$${1 \over 2},2\sqrt 2 $$
$${1 \over 4},2\sqrt 2 $$
$${1 \over 2},\sqrt 2 $$
$${1 \over 4},\sqrt 2 $$
Explanation
$${1 \over {{K_{eq}}}} = {1 \over {{K_1}}} + {1 \over {{K_1}}}$$
$$ \Rightarrow {K_{eq}} = {{{K_1} \times {K_1}} \over {{K_1} + {K_2}}} = {{K_1^2} \over {2{K_1}}} = {{{K_2}} \over 2}$$
$$ \therefore $$ $$T' = 2\pi {\sqrt {{m \over {{K_{eq}}}}} } $$
$$ = 2\pi {\sqrt {{m \over {{{{K_1}} \over 2}}}} } $$
$$ = 2\pi {\sqrt {{{2m} \over {{K_1}}}} } $$
$$ = \sqrt 2 T$$ [ where $$T = 2\pi {\sqrt {{m \over {{K_1}}}} } $$]
$$ \Rightarrow {K_{eq}} = {{{K_1} \times {K_1}} \over {{K_1} + {K_2}}} = {{K_1^2} \over {2{K_1}}} = {{{K_2}} \over 2}$$
$$ \therefore $$ $$T' = 2\pi {\sqrt {{m \over {{K_{eq}}}}} } $$
$$ = 2\pi {\sqrt {{m \over {{{{K_1}} \over 2}}}} } $$
$$ = 2\pi {\sqrt {{{2m} \over {{K_1}}}} } $$
$$ = \sqrt 2 T$$ [ where $$T = 2\pi {\sqrt {{m \over {{K_1}}}} } $$]
Comments (0)
