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The Strouhal number for flow past cylinder is roughly 0.183 as reported by Williamson (link).
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In order to capture the shedding correctly, we should have at least 20 to 25 time steps in one shedding cycle. Let's use 25 for our case.
{latex}
\large
$$
{\rm{Sr}} = {fD \over U} = 0.183
$$
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For our case, *D = 2, U = 1*
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Therefore, shedding frequency *f = 0.0915*
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Cycle time,

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\large
$$
{\rm{t}} = {1 \over f} = 10.9 sec
$$
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*t = 1/f = 10.9 sec*
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Therefore, *Time Step Size = 10.9/25 = 0.436 sec*