The displacement y of a particle at position x and time t is given...

According to the principle of dimensional homogeneity, the arguments of trigonometric functions must be dimensionless. Thus, the quantities $Bt and Cx$ are dimensionless.

$[B][T] = [M^0 L^0 T^0] \Rightarrow [B] = [T^{-1}]$

$[C][L] = [M^0 L^0 T^0] \Rightarrow [C] = [L^{-1}]$

Since $y$ represents displacement, $[A] = [L]$.

The dimensional formula for the required quantity is:
\left[\dfrac{AB}{C}\right] = \dfrac{[L][T^{-1}]}{[L^{-1}]} = [L^2 T^{-1}]$<br><br>Let$[L^2 T^{-1}] = [F]^a [\rho]^b [T]^c$<br><br>Substituting the standard dimensions for Force ($[MLT^{-2}]$), Density ($[ML^{-3}]$), and Time ($[T]$):<br>$[M^0 L^2 T^{-1}] = [MLT^{-2}]^a [ML^{-3}]^b [T]^c$<br>$[M^0 L^2 T^{-1}] = [M^{a+b} L^{a-3b} T^{-2a+c}]$<br><br>Comparing the powers of$M$,$L$, and$T$on both sides:<br>$a + b = 0 \Rightarrow b = -a$<br>$a - 3b = 2$<br>$-2a + c = -1$<br><br>Substituting$b = -a$into the second equation:<br>$a - 3(-a) = 2 \Rightarrow 4a = 2 \Rightarrow a = \dfrac{1}{2}$<br>Thus,$b = -\dfrac{1}{2}$.<br><br>Substituting$a = \dfrac{1}{2}$into the third equation:<br>$-2\left(\dfrac{1}{2}\right) + c = -1 \Rightarrow -1 + c = -1 \Rightarrow c = 0$<br><br>Therefore, the dimensions of$\dfrac{AB}{C}$in the new system are$[F^{1/2} \rho^{-1/2} T^0]$.<br><br>Answer:$[F^{1/2} \rho^{-1/2} T^0]