Q111 / 160 · JIPMER 2013 · Physics

Two simple pendulums of length 0.5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed $x$ oscillations, where $x$ is

  • A
    1
  • B
    3
  • Correct answer
    C
    2
  • D
    5

Solution

Time period of simple pendulum For Ist pendulum $\begin{aligned} T_1 & =2 \pi \sqrt{\frac{5}{g}} \\ & =2 \pi \sqrt{\frac{5}{9.8}}=\sqrt{20} \mathrm{~s}\end{aligned}$ For IInd pendulum $T_2=2 \pi \sqrt{\frac{20}{g}}$ $=2 \pi \sqrt{\frac{20}{9.8}}=2 \sqrt{20} \mathrm{~s}$ Suppose the time will be in same phase is $t$ then $\begin{aligned} \frac{1}{t} & =\frac{1}{T_1}-\frac{1}{T_2} \\ & =\frac{1}{\sqrt{20}}-\frac{1}{2 \sqrt{20}}\end{aligned}$ $=\frac{1}{2 \sqrt{20}}$ So, $\quad t=2 \sqrt{20}$ Hence, the number of oscillation of simple pendulum of shorter length $(x)=\frac{2 \sqrt{20}}{\sqrt{20}}=2$.

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

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