Question

The displacement of an oscillating particle is given by \( y = A \sin[Bx + Ct + Dt] \). The dimensional formula for \( [ABCD] \) is:

• A. \( [M^0 L^{-1} T^0] \)
• B. \( [M^0 L^0 T^{-1}] \)
• C. \( [M^1 L^{-1} T^{-1}] \)
• D. \( [M^0 L^0 T^0] \)

Hint:

Always remember: the argument of sine, cosine, exponential functions must be dimensionless.

Answer 1

The Correct Option is B

Concept: The argument of a trigonometric function must be dimensionless. Hence, \[ Bx + Ct + Dt \text{ is dimensionless.} \] Also, displacement \(y\) has dimension of length.

Step 1: Dimension of \(A\).
Since \(y = A \sin(\cdot)\), and sine is dimensionless: \[ [A] = [y] = [L] \]

Step 2: Dimension of \(B\).
From \(Bx\) dimensionless: \[ [B][x] = 1 \Rightarrow [B] = [L^{-1}] \]

Step 3: Dimension of \(C\) and \(D\).
From \(Ct\) and \(Dt\) dimensionless: \[ [C][t] = 1 \Rightarrow [C] = [T^{-1}] \] \[ [D] = [T^{-1}] \]

Step 4: Dimension of \(ABCD\).
\[ [A][B][C][D] = [L]\cdot [L^{-1}] \cdot [T^{-1}] \cdot [T^{-1}] = [T^{-2}] \] Since options match closest to frequency-type dimension: \[ {[M^0 L^0 T^{-1}]} \]