Respuesta :
ANSWERS
a. f₁ = 380 Hz; f₂ = 760 Hz; f₃ = 1140 Hz
b. f₁ = 190 Hz; f₃ = 570 Hz; f₅ = 950 Hz
EXPLANATION
a. For a pipe of length L open at both ends, the frequencies of the first three harmonics are:
[tex]\begin{cases}f_1=\frac{v}{2L} \\ \\ f_2=2f_1=\frac{v}{L} \\ \\ f_3=3f_1=\frac{3v}{2L}\end{cases}[/tex]Assuming that the speed of the wave is the speed of sound: 343 m/s and knowing that the length of the pipe is L = 45.132 cm = 0.45132 m we can find the frequencies of the first three harmonics:
[tex]\begin{cases}f_1=\frac{343m/s}{2\cdot0.45132m}\approx380Hz \\ \\ f_2=2f_1=2\cdot380Hz\approx760Hz \\ \\ f_3=3f_1=3\cdot380Hz\approx1140Hz\end{cases}[/tex]b. For a pipe of length L closed at one end and open at the other, the frequencies of the first three harmonics are:
[tex]\begin{cases}f_1=\frac{v}{4L} \\ \\ f_2=DNE \\ \\ f_3=3f_1=\frac{3v}{4L}\end{cases}[/tex]In a closed pipe, there can only be odd harmonics (1, 3, 5...). Therefore, the second harmonic does not exist and the "third harmonic" would be the 5th,
[tex]\begin{cases}f_1=\frac{v}{4L} \\ \\ f_3=3f_1=\frac{3v}{4L} \\ \\ f_5=5f_1=\frac{5v}{4L}\end{cases}[/tex]Again, the length of the pipe is 45.132 cm = 0.45132 m, so the first three harmonics are:
[tex]\begin{cases}f_1=\frac{343m/s}{4\cdot0.45132m}\approx190Hz \\ \\ f_3=3f_1=3\cdot190Hz=570Hz \\ \\ f_5=5f_1=5\cdot190Hz=950Hz\end{cases}[/tex]