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In the relation $\frac{dy}{dt} = 2 ωsin (ωt + ϕ_{0})$ the dimensional formula for $ωt + ϕ_{0}$ is

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Correct option is D

Here ωt+ϕ0 is dimensionless because it is an argument of a trigonometric function.

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In the relation dydt=2ωsin⁡ωt+ϕ0the dimensional formula for ωt+ϕ0is

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In the relation $\frac{dy}{dt} = 2 ωsin (ωt + ϕ_{0})$ the dimensional formula for $ωt + ϕ_{0}$ is