The force F is given in terms of time t and displacement x by the equation `F=A cos Bx+Csin Dt`. The dimensions of `(D)/(B)` are

Options

A. (a)`M^(@)L^(@)T^(@)`
B. (b)`M^(@)L^(@)T^(-1)`
C. (c)`M^(@)L^(-1)T^(@)`
D. (d)`M^(@)L^(1)T^(-1)`

Answer

To find the dimensions of \( \frac{D}{B} \) from the given equation \( F = A \cos(Bx) + C \sin(Dt) \), we need to analyze the terms inside the cosine and sine functions.

### Step-by-Step Solution:

1. **Identify the Terms**:
The equation is given as \( F = A \cos(Bx) + C \sin(Dt) \). Here, \( Bx \) and \( Dt \) must be dimensionless because they are arguments of the cosine and sine functions.

2. **Dimensions of Displacement \( x \)**:
The displacement \( x \) has the dimension of length, which is represented as:
\[ [x] = L \]

3. **Dimensions of Time \( t \)**:
The time \( t \) has the dimension of time, represented as:
\[ [t] = T \]

4. **Finding Dimensions of \( B \)**:
Since \( Bx \) is dimensionless, we can write:
\[ [B][x] = 1 \quad \text{(dimensionless)} \]
Therefore, the dimensions of \( B \) can be expressed as:
\[ [B] = \frac{1}{[x]} = \frac{1}{L} = L^{-1} \]

5. **Finding Dimensions of \( D \)**:
Similarly, since \( Dt \) is also dimensionless, we have:
\[ [D][t] = 1 \quad \text{(dimensionless)} \]
Thus, the dimensions of \( D \) are:
\[ [D] = \frac{1}{[t]} = \frac{1}{T} = T^{-1} \]

6. **Finding Dimensions of \( \frac{D}{B} \)**:
Now we can find the dimensions of \( \frac{D}{B} \):
\[ \frac{[D]}{[B]} = \frac{T^{-1}}{L^{-1}} = T^{-1} \cdot L^{1} = \frac{L}{T} \]

### Final Answer:
The dimensions of \( \frac{D}{B} \) are:
\[ \frac{D}{B} = \frac{L}{T} \]