A tuning fork with frequency 800 ~Hz produces resonance in a...

For vibrating tunning fork over a resonance tube, the first resonance is obtained at the length
$$l_1=\frac{\lambda}{4}$$ and for second resonance,
$$\mathrm{l}_2=\frac{\lambda}{4}+\frac{\lambda}{2}=\frac{3 \lambda}{4}$$ From Eq. (i) and (ii), we get \begin{array}{cc} & \mathrm{l}_2-\mathrm{l}_1=\frac{3 \lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2} \\ \Rightarrow & \lambda=2\left(l_2-l_1\right) \\ \Rightarrow & \mathrm{v}=2 \mathrm{f}\left(\mathrm{l}_2-\mathrm{l}_1\right) \end{array} ..(iii) $\left(\because \lambda=\frac{\mathrm{v}}{\mathrm{f}}\right)$ Here, $\mathrm{f}=800 \mathrm{~Hz}, \mathrm{l}_1=9.75 \mathrm{~cm}, \mathrm{l}_2=31.25 \mathrm{~cm}$ Substituting the given values in Eq. (iii), we get \begin{aligned} & \Rightarrow \mathrm{v}=2 \times 800(31.25-9.75) \\ & =34400 \mathrm{~cm} / \mathrm{s}=344 \mathrm{~m} / \mathrm{s} \end{aligned}