Q 11-14-123JEE MainJEE Main 2019 (11 Jan, Shift 1)Easy
Equation of travelling wave on a stretched string of linear density $5$ g/m is $y = 0.03\sin(450t - 9x)$ where distance and time are measured in SI units. The tension in the string is:
Answer: (C) $12.5$ N
Wave speed $v = \dfrac{\omega}{k} = \dfrac{450}{9} = 50$ m/s.
$$T = \mu v^2 = 5\times10^{-3}\times2500 = 12.5\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics