Q153 / 170 · Manipal (MET) 2021 · Physics
The displacement of an oscillating particle is given by $y=A \sin [B x+C t+D t]$. The dimensional formula for $[A B C D]$ is :
- A$\left[\mathrm{M}^0 \mathrm{~L}^{-1} \mathrm{~T}^0\right]$
- Correct answerB$\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$
- C$\left[\mathrm{M}^\rho \mathrm{L}^{-1} \mathrm{~T}^{-1}\right]$
- D$\left[M^0 L^0 T^0\right]$
Solution
$y=A \sin [B x+C t+D]$
As each term inside the bracket is dimension less, $\therefore \quad[A]=[y]=[L], \quad[B]=\frac{1}{[x]}=\left[L^{-1}\right]$
$[C]=\frac{1}{[t]}=\left[1^{-1}\right]$ and $D$ is dimensionless. $\therefore[A B C D]=[\mathrm{L}]\left[\mathrm{L}^{-1}\right]\left[\mathrm{T}^{-1}\right][1]=\left[\mathrm{N}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$
As each term inside the bracket is dimension less, $\therefore \quad[A]=[y]=[L], \quad[B]=\frac{1}{[x]}=\left[L^{-1}\right]$
$[C]=\frac{1}{[t]}=\left[1^{-1}\right]$ and $D$ is dimensionless. $\therefore[A B C D]=[\mathrm{L}]\left[\mathrm{L}^{-1}\right]\left[\mathrm{T}^{-1}\right][1]=\left[\mathrm{N}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$
Hence, the correct answer is option: 'B'.