If the displacement y of a particle is y =A sin (pt+qx) then dimensional formula of pq is

Options

A. L
B. `LT^(-1)`
C. `T^(-1)`
D. `L^(-1) T^(-1)`

Answer

To find the dimensional formula of the product \( pq \) given the displacement \( y = A \sin(pt + qx) \), we need to ensure that the arguments of the sine function are dimensionless. This means that both \( pt \) and \( qx \) must be dimensionless.

### Step 1: Analyze the term \( pt \)
The term \( pt \) must be dimensionless. Therefore, we can write:
\[ [pt] = [p][t] \]
Since time \( t \) has the dimension \( [T] \), we have:
\[ [pt] = [p][T] = [M^0 L^0 T^1] \]
This implies:
\[ [p][T] = [M^0 L^0 T^0] \]
From this, we can deduce that:
\[ [p] = [M^0 L^0 T^{-1}] = T^{-1} \]

### Step 2: Analyze the term \( qx \)
Similarly, for the term \( qx \) to be dimensionless, we can write:
\[ [qx] = [q][x] \]
Since distance \( x \) has the dimension \( [L] \), we have:
\[ [qx] = [q][L] = [M^0 L^1 T^0] \]
This implies:
\[ [q][L] = [M^0 L^0 T^0] \]
From this, we can deduce that:
\[ [q] = [M^0 L^{-1} T^0] = L^{-1} \]

### Step 3: Find the dimensional formula of \( pq \)
Now, we can find the dimensional formula of the product \( pq \):
\[ [pq] = [p][q] = [M^0 L^0 T^{-1}][M^0 L^{-1} T^0] \]
Calculating this gives:
\[ [pq] = [M^0 L^{-1} T^{-1}] \]

### Final Answer
Thus, the dimensional formula of \( pq \) is:
\[ [M^0 L^{-1} T^{-1}] \]