Final answer:
To find the dimension of d/b in the equation F = Acos(bx) + Csin(dt), we analyze the dimension of d as [T^{-1}] because it is associated with time in a trigonometric function. Since b must also be dimensionless, the final dimension of d/b is [T].
Step-by-step explanation:
If F = Acos(bx) + Csin(dt), then to find the dimension of d/b, we need to ensure that both terms on the right side of the equation have the same dimensions as the force, F, on the left side. Since force has dimensions of [F] = MLT‑2, and the dimensions of cos(bx) and sin(dt) are dimensionless, A must have dimensions of force ([A] = MLT‑2), and so must C ([C] = MLT‑2). Looking at the term Csin(dt), we know that d must have dimensions of T‑1 because it is multiplied by time, t, which has dimensions of [t] = T. Hence, the dimension of d is [T‑1]. And since b is an equivalent term used in a trigonometric function, it must also be dimensionless, which is consistent with being the inverse of a time as well. Therefore, the dimension of d/b is indeed [T].