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

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(D) The given equation is $F = A \cos(Bx) + C \sin(Dt).$According to the principle of homogeneity of dimensions,the argument of any trigonometric func

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$M^0 L^1 T^{-1}$

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(D) The given equation is $F = A \cos(Bx) + C \sin(Dt).$According to the principle of homogeneity of dimensions,the argument of any trigonometric function must be dimensionless.Therefore,the dimensions of $Bx$ must be $[M^0 L^0 T^0]$.$[B] \cdot [x] = [M^0 L^0 T^0]$$[B] \cdot [L] = [M^0 L^0 T^0]$$[B] = [L^{-1}]$Similarly,the dimensions of $Dt$ must be $[M^0 L^0 T^0]$.$[D] \cdot [t] = [M^0 L^0 T^0]$$[D] \cdot [T] = [M^0 L^0 T^0]$$[D] = [T^{-1}]$Now,the dimensional formula of $D/B$ is:$[D/B] = \frac{[T^{-1}]}{[L^{-1}]} = [L^1 T^{-1}]$

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 $D/B$ is

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