242.A wire of length 1m and mass 20g is stretched with a force of 800N.The frequencies of first two overtones are

Dear Student,

Please find below the solution to the asked query:

The equation for the frequency of vibrations on a stretched string is,

$$\begin{gathered} f_n=\frac{n}{2l}\sqrt{\frac{T}{\mu }}\Rightarrow f_n=\frac{n}{2\times 1}\sqrt{\frac{800}{{\displaystyle \raisebox{1ex}{$20\times {10}^{-3}$}\!\left/ \!\raisebox{-1ex}{$1$}\right.}}} \\ \Rightarrow f_n=\frac{n}{2}\sqrt{40000}\Rightarrow f_n=100n \\ For first overtone, \\ {f}_2=100\times 2=200 Hz, \\ For second overtone, \\ {f}_3=100\times 3=300 Hz \end{gathered}$$

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