In the relation: frac{dy}{dx} = 2omega sin(om | vedclass

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Question

(D) In the given trigonometric function $\sin(\omega t + \phi_0)$,the argument of the sine function,which is $(\omega t + \phi_0)$,must be a dimension

Vedclass · Accepted Answer

$M^0L^0T^0$

Vedclass · Answer

(D) In the given trigonometric function $\sin(\omega t + \phi_0)$,the argument of the sine function,which is $(\omega t + \phi_0)$,must be a dimensionless quantity.This is because trigonometric functions are defined only for dimensionless angles (radians).Therefore,the dimensions of $(\omega t + \phi_0)$ are $[M^0L^0T^0]$,which represents a dimensionless quantity.Thus,the correct option is $(d)$.

In the relation: $\frac{dy}{dx} = 2\omega \sin(\omega t + \phi_0)$,the dimensional formula for $(\omega t + \phi_0)$ is:

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