Two simple pendulums whose lengths are 100 cm and 121 cm are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum, will the two be in phase again?

Let T₁  and T₂ be the time period of the two pendulums,

${\mathrm{T}}_1=2\mathrm{\pi }\sqrt{\frac{100}{\mathrm{g}}}$ and ${\mathrm{T}}_2=2\mathrm{\pi }\sqrt{\frac{121}{\mathrm{g}}}$         (T₁ < T₂ because l₁ < l₂)

Lei at t = 0, they start swinging together. Since, their time periods arc different. the swinging will not be same always.

Only when number of completed oscillation differs by an integer, the two pendulum will again begin to swing together.
Let longer length pendulum complete n oscillation and shorter length pendulum complete (n + 1) oscillation, for the two be in same phase, then (n + 1)T₁ = nT₂.

   $(\mathrm{n}+1)\times 2\mathrm{\pi }\sqrt{\frac{100}{\mathrm{g}}}=\mathrm{n}\times 2\mathrm{\pi }\sqrt{\frac{121}{\mathrm{g}}}$

$\Rightarrow$                         n = 10