Q141 / 150 · MHT CET 2021 (23 Sep Shift 1) · Physics

Two identical springs of constant ' $\mathrm{K}$ ' are connected in series and parallel in shown in figure. A mass ' $\mathrm{M}$ ' is suspended from them. The ratio of their frequencies is series to parallel combination will be Figure

  • Correct answer
    A
    1:2
  • B
    1:4
  • C
    4:1
  • D
    $1: \sqrt{2}$

Solution

In series combination, the effective spring constant $$ \mathrm{k}_1=\frac{\mathrm{k}}{2} $$ In the parallel combination, the effective spring constant is $$ \mathrm{k}_2=2 \mathrm{k} $$ $$ \begin{aligned} & \mathrm{f}_1=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_1}{\mathrm{~m}}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{2 \mathrm{~m}}} \\ & \mathrm{f}_2=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_2}{\mathrm{~m}}}=\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}} \\ & \therefore \frac{\mathrm{f}_1}{\mathrm{f}_2}=\sqrt{\frac{1}{4}}=\frac{1}{2} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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