Given that the displacement of an oscillating particle is given by `y=Asin(Bx+Ct+D)`. The dimensional formula for `(ABCD)` is

Options

A. `[M^(0)L^(-1)T^(0)]`
B. `[M^(0)L^(0)T^(-1)]`
C. `[M^(0)L^(-1)T^(-1)]`
D. `[M^(0)L^(0)T^(0)]`

Answer

To find the dimensional formula for the product \( ABCD \) given the displacement of an oscillating particle \( y = A \sin(Bx + Ct + D) \), we need to analyze the dimensions of each variable involved.

### Step 1: Understand the displacement equation
The equation \( y = A \sin(Bx + Ct + D) \) indicates that \( y \) is a displacement, which has the dimension of length. Therefore, we can write:
\[ [Y] = [A] \]
where \( [Y] \) is the dimension of displacement.

### Step 2: Determine the dimension of \( A \)
Since \( y \) is a displacement, we have:
\[ [A] = [L] \]
where \( [L] \) represents the dimension of length.

### Step 3: Analyze the term \( Bx \)
The term \( Bx \) must be dimensionless because it is an argument of the sine function. Therefore, we can express this as:
\[ [B][X] = [1] \]
where \( [X] \) is the dimension of \( x \) (which is length, \( [L] \)). Thus:
\[ [B][L] = [1] \implies [B] = [L^{-1}] \]

### Step 4: Analyze the term \( Ct \)
Similarly, for the term \( Ct \) to be dimensionless, we have:
\[ [C][T] = [1] \]
where \( [T] \) is the dimension of time. Thus:
\[ [C][T] = [1] \implies [C] = [T^{-1}] \]

### Step 5: Analyze the term \( D \)
The term \( D \) is also dimensionless, so we have:
\[ [D] = [1] \]

### Step 6: Combine the dimensions of \( A \), \( B \), \( C \), and \( D \)
Now we can find the dimensional formula for the product \( ABCD \):
\[ [ABCD] = [A][B][C][D] = [L][L^{-1}][T^{-1}][1] \]

### Step 7: Simplify the dimensions
Combining these dimensions gives:
\[ [ABCD] = [L][L^{-1}][T^{-1}] = [T^{-1}] \]

### Final Result
Thus, the dimensional formula for \( ABCD \) is:
\[ \boxed{[M^0 L^0 T^{-1}]} \]