Q122 / 160 · TS EAMCET 2024 (09 May Shift 1) · Physics

The equation of motion of a damped oscillator is given by $m \frac{d^2 x}{d t^2}+b \frac{d x}{d t}+k x=0$. The dimensional formula of $\frac{b}{\sqrt{k m}}$ is

  • Correct answer
    A
    $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$
  • B
    $\left[\mathrm{M}^0 \mathrm{~L}^{\mathrm{L}} \mathrm{T}^{-2}\right]$
  • C
    $\left[M^1 L^1 T^{-2}\right]$
  • D
    $\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$

Solution

From equation given $\frac{\mathrm{md}^2 \mathrm{x}}{\mathrm{dt}{ }^2}+\mathrm{b} \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{kx}=0$ $\mathrm{F}=\mathrm{kx} \Rightarrow \mathrm{k}=\frac{\mathrm{F}}{\mathrm{x}} \text { and } \mathrm{b} \frac{\mathrm{dx}}{\mathrm{~d} t}=\mathrm{F} \Rightarrow \mathrm{~b}=\frac{\mathrm{F}}{\mathrm{dx} / \mathrm{dt}}=\frac{\mathrm{F}}{\mathrm{~V}}$
Thereforedimensions of $\frac{\mathrm{b}}{\sqrt{\mathrm{km}}}=\frac{\mathrm{F} / \mathrm{V}}{\sqrt{\frac{\mathrm{F}}{\mathrm{X}} \mathrm{m}}}=\frac{\mathrm{MLT}^{-2} / \mathrm{LT}^{-1}}{\sqrt{\frac{\mathrm{MLT}^{-2} \cdot \mathrm{M}}{\mathrm{L}}}}$ $=\frac{\mathrm{MT}^{-1}}{\mathrm{MT}^{-1}}=\left[\mathrm{M}^{0} \mathrm{L}^{0} \mathrm{T}^{0}\right]$

Hence, the correct answer is option: 'A'.

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