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Two simple pendulums of length $0.5\, m$ and $2.0\, m$ 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 .... oscillations.
$5$
$1$
$2$
$3$
Solution
(c) If $t$ is the time taken by pendulums to come in same phase again first time after $t = 0$.
and ${N_S} = $ Number of oscillations made by shorter length pendulum with time period ${T_S}$.
${N_L} = $ Number of oscillations made by longer length pendulum with time period ${T_L}$.
Then $t = {N_S}{T_S} = {N_L}{T_L}$
$ \Rightarrow {N_S}2\pi \sqrt {\frac{5}{g}} = {N_L} \times 2\pi \sqrt {\frac{{20}}{g}} $ ($\because T = 2\pi \sqrt {\frac{l}{g}} $)
$ \Rightarrow {N_S} = 2{N_L}$ i.e. if ${N_L} = 1$
$ \Rightarrow {N_S} = 2$
