91Ó°ÊÓ

The fundamental frequency is the minimum frequency of periodic sound waveform.

Mathematically, speed of a wave is

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

Where F is the tension force and µ is the mass per length value.

$\mathbf{v}\mathbf{=}\sqrt{\frac{\mathbf{FL}}{\mathbf{M}}}$

For fundamental frequency,

$$\begin{gathered} \mathbf{f}\mathbf{=}\frac{\mathbf{v}}{\mathbf{λ}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{λ}\mathbf{=}\mathbf{2}\mathbf{λ} \\ \mathbf{So}\mathbf{,}\mathbf{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\sqrt{\frac{\mathbf{F}}{\mathbf{mL}}} \end{gathered}$$

Use differential evaluation to determine the small change in frequency corresponding to the small change in length due to thermal expansion.

$$\begin{gathered} ∆\mathrm{f}≈\frac{∂\mathrm{f}}{∂\mathrm{L}}∆\mathrm{L},\mathrm{where}∆\mathrm{L}=\mathrm{³¢Î\pm }∆\mathrm{T}\mathrm{is}\mathrm{the}\mathrm{thermal}\mathrm{expansion}\mathrm{in}\mathrm{length} \\ ∆\mathrm{f}=\frac{1}{2}\frac{1}{2}{\left(\frac{\mathrm{F}}{\mathrm{mL}}\right)}^{\frac{1}{2}}\left(\frac{\mathrm{F}}{\mathrm{m}}\right)\left(-\frac{1}{{\mathrm{L}}^2}\right)∆\mathrm{L} \\ =-\frac{1}{2}\left(\frac{1}{2}\sqrt{\frac{\mathrm{F}}{\mathrm{mL}}}\right)\frac{∆\mathrm{L}}{\mathrm{L}}=\frac{1}{2}\mathrm{f}\frac{∆\mathrm{L}}{\mathrm{L}} \\ ∆\mathrm{f}=-\frac{1}{2}\left(\mathrm{Î\pm }∆\mathrm{T}\right)\mathrm{f} \\ =-\frac{1}{2}(1.7×{10}^{-5}{\left(°\mathrm{C}\right)}^{-1})\left(40°\mathrm{C}\right)\left(440\mathrm{Hz}\right) \\ =-0.15\mathrm{Hz} \end{gathered}$$

Thus, the frequency of fundamental note decreases as length increases.

Similarly, use differential evaluation to determine the change in speed,

$$\begin{gathered} ∆\mathrm{v}≈\frac{∂\mathrm{v}}{∂\mathrm{L}}∆\mathrm{L} \\ \frac{∆\mathrm{v}}{\mathrm{v}}=\frac{{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{\mathrm{FL}}{\mathrm{m}}}\right)}^{-{\displaystyle \frac{1}{2}}}\left({\displaystyle \frac{\mathrm{F}}{\mathrm{m}}}\right)∆\mathrm{L}}{\sqrt{{\displaystyle \frac{\mathrm{FL}}{\mathrm{m}}}}}=\frac{∆\mathrm{L}}{2\mathrm{L}}=\frac{\mathrm{Î\pm }∆\mathrm{T}}{2} \\ =\frac{1}{2}\left(1.7×{10}^{-5}{\left(°\mathrm{C}\right)}^{-1}\right)\left(40°\mathrm{C}\right) \\ =3.4×{10}^{-4} \\ =0.034\% \end{gathered}$$

Thus, the percentage increase in speed of wave is 0.034%

Substitute all the values,

$\frac{∆\mathrm{λ}}{\mathrm{λ}}=\left(1.7×{10}^{-5}{\left(°\mathrm{C}\right)}^{-1}\right)\left(40°\mathrm{C}\right)=6.8×{10}^{-4}=0.068\%$

Thus the percentage increase in wavelength with respect to increase in length is 0.068%.