Question
The force $F$ is given in terms of time $t$ and displacement $x$ by the equation $F=A \cos B x+C \sin D t$. The dimensions of $\frac{D}{B}$ are(A) $M^{0} L^{0} T^{0}$(B) $M^{0} L^{0} T^{-1}$(C) $M^{0} L^{-1} T^{0}$(D) $M^{0} L^{1} T^{-1}$
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Therefore, the dimensions of $Bx$ and $Dt$ should be the same and dimensionless. Show more…
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Force $F$ is given in terms of time $t$ and distance $x$ by $F=A \sin C t+B \cos D x$. Then, dimensions of $A / B$ and $C / D$ are a. $\left[M^{0} L^{0} T^{0}\right],\left[M^{0} L^{0} T^{-1}\right]$ b. $\left[M L T^{-2}\right],\left[M^{0} L^{-1} T^{0}\right]$ c. $\left[M^{0} L^{0} T^{0}\right],\left[M^{0} L T^{-1}\right]$ d. $\left[M^{0} L^{\prime} T^{\prime-1}\right],\left[M^{0} L^{0} T^{0}\right]$
The force $F$ is given in terms of time $t$ and displacement $x$ by the equation: $$ F=a \cos \alpha x+b \sin \beta t $$ where $a$ and $b$ are the amplitudes. The dimensions of $\beta / \alpha$ are: (a) $\left[M^{0} L^{\circ} T^{0}\right]$ (b) $\left[M^{0} L^{0} T^{-1}\right]$ (c) $\left[M^{0} L^{-1} T^{0}\right]$ (d) $\left[M^{\circ} L^{1} T^{-1}\right]$
A force $F$ is given by $F=a t+b t^{2}$ where $t$ is time, what are the dimension of $a$ and $b ?$ (A) $\left[M L T^{-1}, M L T^{0}\right]$, (B) $\left[M L T^{-3}, M L^{2} T^{4}\right]$ (C) $\left[M L T^{-4}, M L T^{-1}\right]$ (D) $\left[M L^{-3} T, M L T^{-4}\right]$
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