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Sagot :
[tex]\begin{gathered} L=0.44\text{ m} \\ f=389.77\text{ Hz} \end{gathered}[/tex]
Explanation
Step 1
Let
[tex]\begin{gathered} \text{fundamental wavelength=0.88 ,} \\ \lambda=2L \\ \text{hence, replace} \\ 0.88m=2\text{ L} \\ \text{divide both sides by 2} \\ \frac{0.88m}{2}=\frac{2\text{ L}}{2} \\ 0.44=L \end{gathered}[/tex]hence, the length of the bottle is 0.44 meters
Step 2
What is the frequency of the third harmonic of this bottle?
For the third harmonic, the length of the string is equivalent to three-halves of a wavelength,so
[tex]\begin{gathered} third\text{ armonci= }\frac{\text{3}}{2}\cdot0.88\text{ m} \\ third\text{ armonci= }1.32\text{ m} \\ \end{gathered}[/tex]so
[tex]\begin{gathered} 1.32=2L \\ \end{gathered}[/tex]replace in the quation and solve for
Let speed of sound = 343 m/s
[tex]\begin{gathered} f=\frac{v}{\lambda} \\ \text{replacing} \\ f=\frac{343}{0.88} \\ f=389.77\text{ Hz} \end{gathered}[/tex]I hope this helps
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