Q110 / 180 · JIPMER 2019 · Physics

Find the distance between two consecutive nodes, if for a string $\mathrm{T}=0.25 \mathrm{~N}, \mu=1 \mathrm{mg} / \mathrm{m} \mathrm{f}$ $=250 \mathrm{~Hz}$.

  • A
    2 m
  • Correct answer
    B
    1 m
  • C
    3 m
  • D
    1.5 m

Solution

Given, Tension in the string, $\mathrm{T}=0.25 \mathrm{~N}$ a mass per unit length, $\mu=1 \mathrm{mg} / \mathrm{m}=10^{-6} \mathrm{~kg} / \mathrm{m}$ Velocity of wave, $\quad v=\sqrt{\frac{T}{H}}\ldots(i)$ The distance between two consecutive nodes is equal to half the wave length ' $\lambda$ '. $\therefore \quad v=f \lambda\ldots(ii)$ Form Eqs. (i) and (ii), $\begin{aligned} \mathrm{f} \lambda=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \Rightarrow \lambda & =\frac{1}{\mathrm{f}} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{250} \sqrt{\frac{0.25}{10^{-6}}} \\ \lambda & =2 \end{aligned}$ $\therefore$ Difference between two consecutive nodes $=\frac{\lambda}{2}=\frac{2}{2}=1 \mathrm{~m}$

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

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