Q 11-14-082JEE MainJEE Main 2021 (31 Aug, Shift 1)Easy
A wire having a linear mass density $9.0\times10^{-4}\ \text{kg m}^{-1}$ is stretched between two rigid supports with a tension of $900$ N. The wire resonates at a frequency of $500$ Hz. The next higher frequency at which the same wire resonates is $550$ Hz. The length of the wire is ______ m.
Numerical value type. Enter your answer.
Answer: 10
Wave speed $v = \sqrt{\dfrac T\mu} = \sqrt{\dfrac{900}{9\times10^{-4}}} = 1000\ \text{m s}^{-1}$.
Successive harmonics differ by the fundamental: $\dfrac{v}{2L} = 550 - 500 = 50$ Hz.
$$L = \frac{1000}{2\times50} = 10\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics