The force F is given in terms of time t and d | vedclass

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Question

(A) The argument of trigonometric functions like $\cos$ and $\sin$ must be dimensionless.For $\cos(Bx)$,the term $Bx$ must be dimensionless,so $[B] =

Vedclass · Accepted Answer

$[ML^{2}T^{-3}]$

Vedclass · Answer

(A) The argument of trigonometric functions like $\cos$ and $\sin$ must be dimensionless.For $\cos(Bx)$,the term $Bx$ must be dimensionless,so $[B] = [x]^{-1} = [L^{-1}]$.For $\sin(Dt)$,the term $Dt$ must be dimensionless,so $[D] = [t]^{-1} = [T^{-1}]$.Since $F = A \cos(Bx) + C \sin(Dt)$,the dimensions of $A$ must be equal to the dimensions of force $F$,so $[A] = [MLT^{-2}]$.Now,we calculate the dimensions of $\frac{AD}{B}$:$[\frac{AD}{B}] = \frac{[A][D]}{[B]} = \frac{[MLT^{-2}][T^{-1}]}{[L^{-1}]}$.$[\frac{AD}{B}] = [MLT^{-3}][L] = [ML^{2}T^{-3}]$.

The force $F$ is given in terms of time $t$ and displacement $x$ by the equation $F = A \cos(Bx) + C \sin(Dt)$. The dimensional formula of $\frac{AD}{B}$ is -

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