Q 11-14-112JEE MainJEE Main 2019 (10 Jan, Shift 1)Medium
A string of length $1\ \text{m}$ and mass $5\ \text{g}$ is fixed at both ends. The tension in the string is $8.0\ \text{N}$. The string is set into vibration using an external vibrator of frequency $100\ \text{Hz}$. The separation between successive nodes on the string is close to
Answer: (A) $20.0\ \text{cm}$
Linear density $\mu = \dfrac{5\times10^{-3}}{1} = 5\times10^{-3}\ \text{kg/m}$:
$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{8}{5\times10^{-3}}} = 40\ \text{m/s},\qquad \lambda = \frac vf = \frac{40}{100} = 0.4\ \text{m}$$
Successive nodes are $\lambda/2 = 20\ \text{cm}$ apart.
Solution by Sreeraj P, M.Sc Physics