Verified Solution • 2025–26 Session moderate

QJEEM21Ph1f07: JEE Main 2021 Physics - Class 11 JEEM21Ph1f07 Solution: If two similar springs each of spring constant K 1 are joined…

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 :

Multiple Choice Options

  • A 12,22{1 \over 2},2\sqrt 2
  • B 14,22{1 \over 4},2\sqrt 2
  • C 12,2{1 \over 2},\sqrt 2
  • D 14,2{1 \over 4},\sqrt 2

Step-by-Step Verified Solution

1Keq=1K1+1K1{1 \over {{K_{eq}}}} = {1 \over {{K_1}}} + {1 \over {{K_1}}}

⇒Keq=K1×K1K1+K2=K122K1=K22\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′=2πmKeqT' = 2\pi {\sqrt {{m \over {{K_{eq}}}}} }

=2πmK12= 2\pi {\sqrt {{m \over {{{{K_1}} \over 2}}}} }

=2π2mK1= 2\pi {\sqrt {{{2m} \over {{K_1}}}} }

=2T= \sqrt 2 T [ where T=2πmK1T = 2\pi {\sqrt {{m \over {{K_1}}}} } ]

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