Animated Solution for Physics - Physics and Measurement: The force 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

When we encounter a physical equation like $F=A\cos (Bx)+C\sin (Dt)$, it might look intimidating at first glance. However, the Principle of Dimensional Homogeneity acts as our ultimate decoding tool. This principle states two fundamental rules that govern all valid physics equations.

First, you can only add or subtract quantities that share the exact same dimensions. You cannot add a mass to a velocity, just as you cannot add apples to oranges. Therefore, the terms $A\cos (Bx)$ and $C\sin (Dt)$ must both independently have the same dimensions as the force $F$.

Second, the arguments of transcendental functions—such as trigonometric, exponential, and logarithmic functions—must be strictly dimensionless. They represent pure numbers or ratios, like angles measured in radians.

Let's apply this logic to the arguments of our trigonometric functions. We start with the term $Bx$. Since $x$ represents displacement, its dimensional formula is simply $[L]$. For the entire product $Bx$ to be a dimensionless angle, the dimension of $B$ must perfectly cancel out the dimension of $x$.

We apply the exact same reasoning to the term $Dt$. Here, $t$ represents time, which has the dimension $[T]$. For $Dt$ to be dimensionless, $D$ must be the reciprocal of time.

Now, let's determine the dimension of the coefficient $A$. Look at the first term of our equation: $A\cos (Bx)$. We already established that the cosine function evaluates to a pure, dimensionless number.

Because the entire term must have the dimensions of force to satisfy homogeneity, the coefficient $A$ must carry the entire dimensional burden of the force. We know from Newton's Second Law that force is mass times acceleration, giving it the dimension $[ML{T}^{-2}]$.

We have successfully gathered all the necessary dimensional pieces: $[A]=[ML{T}^{-2}]$, $[B]=[L^{-1}]$, and $[D]=[T^{-1}]$. Our final objective is to find the dimensional formula for the expression $\frac{AD}{B}$.

Let's substitute our findings into the target expression:

Now, we carefully apply the laws of exponents to simplify the expression. In the numerator, the time dimensions combine: $T^{-2}\cdot T^{-1}=T^{-3}$. The length dimension in the denominator, $L^{-1}$, moves to the numerator as $L^1$, multiplying with the existing $L$ to become $L^2$.

This elegant process of dimensional analysis not only solves the problem but also reveals the underlying structural symmetry of the physical equation.