Question

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

• A. $ \left[ {{M}^{0}}{{L}^{-1}}{{T}^{0}} \right] $
• B. $ \left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right] $
• C. $ \left[ {{M}^{0}}{{L}^{-1}}{{T}^{-1}} \right] $
• D. $ \left[ {{M}^{0}}{{L}^{0}}{{T}^{0}} \right] $

Answer 1

Answer: (B)

$ y=A\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}}] $