Force `F` is given in terms of time `t` and distance `x` by `F = A sin Ct + B cos Dx`. Then the dimensions of `A//B` and `C//D` are

Options

A. `["M"^(0)"L"^(0)"T"^(0)],["M"^(0)"LT"^(-1)]`
B. `["MLT"^(-2)],["M"^(0)"L"^(0)"T"^(-1)]`
C. `["MLT"^(-2)],["M"^(0)"L"^(-1)"T"^(0)]`
D. `["M"^(0)"LT"^(-1)],["M"^(0)"L"^(0)"T"^(0)]`

Explanation

To solve the problem, we need to analyze the given equation for force \( F \) in terms of time \( t \) and distance \( x \):

\[ F = A \sin(Ct) + B \cos(Dx) \]

### Step 1: Understand the dimensions of the trigonometric functions
Trigonometric functions like sine and cosine are dimensionless. Therefore, the arguments of these functions must also be dimensionless. This leads us to the following equations:

1. The argument of \( \sin(Ct) \) must be dimensionless:
\[ [C][t] = 1 \]
From this, we can deduce the dimensions of \( C \):
\[ [C] = [t]^{-1} = T^{-1} \]

2. The argument of \( \cos(Dx) \) must also be dimensionless:
\[ [D][x] = 1 \]
From this, we can deduce the dimensions of \( D \):
\[ [D] = [x]^{-1} = L^{-1} \]

### Step 2: Analyze the dimensions of \( A \) and \( B \)
Since \( F \) is a force, its dimensions are:
\[ [F] = MLT^{-2} \]

Now, since \( A \sin(Ct) \) and \( B \cos(Dx) \) must have the same dimensions as \( F \), we can write:
\[ [A \sin(Ct)] = [A] \cdot 1 = [A] \]
\[ [B \cos(Dx)] = [B] \cdot 1 = [B] \]

Thus, we have:
\[ [A] = [F] = MLT^{-2} \]
\[ [B] = [F] = MLT^{-2} \]

### Step 3: Find the dimensions of \( \frac{A}{B} \)
Now, we can find the dimensions of \( \frac{A}{B} \):
\[ \frac{[A]}{[B]} = \frac{MLT^{-2}}{MLT^{-2}} = 1 \]
This means:
\[ \frac{A}{B} \text{ is dimensionless.} \]

### Step 4: Find the dimensions of \( \frac{C}{D} \)
Now we can find the dimensions of \( \frac{C}{D} \):
\[ \frac{[C]}{[D]} = \frac{T^{-1}}{L^{-1}} = T^{-1} \cdot L^{1} = \frac{L}{T} \]
Thus, the dimensions of \( \frac{C}{D} \) are:
\[ \frac{C}{D} = \frac{L}{T} \]

### Summary of Results
1. The dimensions of \( \frac{A}{B} \) are dimensionless (\( M^0L^0T^0 \)).
2. The dimensions of \( \frac{C}{D} \) are \( M^0L^1T^{-1} \).