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
Concept. $f_1 = \dfrac{v}{2L}$ with $v = \sqrt{\dfrac{T}{\mu}}$.
Step 1. Linear density. $\mu = \dfrac{0.005}{0.5} = 0.01$ kg/m.
Step 2. Wave speed. $v = \sqrt{\dfrac{400}{0.01}} = 200$ m/s.
Step 3. Fundamental. $f_1 = \dfrac{200}{2 \times 0.5} = 200$ Hz.
Answer. $200$.
Solution by Sreeraj P, M.Sc Physics