Q 11-14-029JEE MainMedium
A wire of length $50$ cm and mass $5$ g is stretched with a tension of $400$ N between two fixed supports. Find its fundamental frequency, in Hz.
Numerical value type. Enter your answer.
Answer: 200
$\mu = \dfrac{0.005}{0.5} = 0.01$ kg/m, $v = \sqrt{\dfrac{400}{0.01}} = 200$ m/s.
$$f_1 = \frac{v}{2L} = \frac{200}{1} = 200\ \text{Hz}$$
Solution by Sreeraj P, M.Sc Physics