Question

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

• 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 1

Answer: (B)

$[M{{\,}^{0}}{{L}^{0}}{{T}^{-1}}]$

Answer 2

$y=A\text{ }sin(Bx+Ct+D)$ As each term inside the bracket is dimensionless, so $A=y=[{{L}^{1}}]$ $B=\frac{1}{x}=[{{L}^{-1}}]$ $C=\frac{1}{t}=[{{T}^{-1}}]$ and D is dimensionless. $\therefore$ $[ABCD]=[L][{{L}^{-1}}][{{T}^{-1}}][1]$ $=[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]$