91Ó°ÊÓ

Length of wire: $L=62.0\text{ c³¾}$

Mass of wire: $\mathrm{m}=7.25\text{ g}$

Tension in wire: $\mathrm{F}=4110\text{ N}$

The sound travels with different velocities in different mediums, depending on their characteristics. For a stretched string, these characteristics are the strig’s tension and mass. Intuition suggests that increasing the tension should increase the restoring forces that tend to straighten the rope when it is disturbed and thus increase the wave speed the equation supports this intuition. The speed of a transverse wave $\left(\mathbf{v}\right)$ in a rope under tension is equal to the square root of the tension$\left(\mathbf{F}\right)$ , divided by the mass per unit length$\left(\mathbf{μ}\right)$ and is given as-

$\mathbf{v}\mathbf{=}\sqrt{\frac{F}{\mathrm{μ}}}$

For the string,

$\begin{array}{c}\mathrm{μ}=\frac{\mathrm{m}}{\mathrm{L}}\\ =\frac{7.25×{10}^{−3}\text{ k²µ}}{0.62\text{″¾}}\\ =0.0117\text{ k²µ/m}\end{array}$

Further calculating the velocity of sound-

$\begin{array}{c}\mathrm{v}=\sqrt{\frac{\mathrm{F}}{\mathrm{μ}}}\\ =\sqrt{\frac{4110\text{ N}}{0.0117\text{ k²µ/m}}}\\ =592.85\text{″¾/s}\end{array}$

For the second overtone, we need to consider the third harmonic$(n=3)$

For the pipe, the frequency is the same $1434.32\text{ H³ú}$

$\begin{array}{c}{\mathrm{f}}_{\mathrm{s}}=\frac{344\text{ }\mathrm{m}/\mathrm{s}}{4×\mathrm{L}}\\ \mathrm{L}=\frac{344\text{ }\mathrm{m}/\mathrm{s}}{4×{\mathrm{f}}_{\mathrm{s}}}\\ =\frac{344\text{ }\mathrm{m}/\mathrm{s}}{4×1434.32\text{ H³ú}}\\ =0.06\text{ }\mathrm{m}\end{array}$

Therefore, the length of the pipe is $6\text{ c³¾}$.