In the equation frac{dy}{dt} = 2omega sin(ome | vedclass

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(D) The given equation is $\frac{dy}{dt} = 2\omega \sin(\omega t + 	heta_0)$.In any trigonometric function,the argument must be a dimensionless quant

Vedclass · Accepted Answer

$M^{0}L^{0}T^{0}$

Vedclass · Answer

(D) The given equation is $\frac{dy}{dt} = 2\omega \sin(\omega t + 	heta_0)$.In any trigonometric function,the argument must be a dimensionless quantity.Since $\sin(\omega t + 	heta_0)$ is a trigonometric function,the argument $(\omega t + 	heta_0)$ must be dimensionless.The dimensions of a dimensionless quantity are given by $[M^{0}L^{0}T^{0}]$.Therefore,the correct option is $D$.

In the equation $\frac{dy}{dt} = 2\omega \sin(\omega t + \theta_0)$,what are the dimensions of $(\omega t + \theta_0)$?

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