Respuesta :
Answer:
The answers to the question are
(a) The integer n that describes the harmonic whose frequency is 571 Hz is 5
(b) The length of the tube is 0.75 m.
B. The depth of the well is 7.12 m.
Explanation:
The nth harmonic of a tube with one end closed is given by
[tex]f_n = n(\frac{v}{4L})[/tex] and the next harmonic is
[tex]f_{n+2} =(n+2)(\frac{v}{4L} )[/tex]
Therefore ividing the first two equations by each other we have
[tex]\frac{f_{n+2}}{f_n} = \frac{n+2}{n}[/tex] from which n = [tex]\frac{2}{\frac{f_{n+2}}{f_n} -1}[/tex]
Therefore the interger that descries the first harmonic is
n = [tex]\frac{2}{\frac{799}{571} -1}[/tex] = 5
(b) We have the length of the pipe given by
L = 5λ/4 but λ = v/f = 343/571 = 0.6 ∴ L = 5*0.6/4 = 0.75 m
B. From n = [tex]\frac{2}{\frac{f_{n+2}}{f_n} -1}[/tex] we have
n for the 36.1 Hz frequency
n = [tex]\frac{2}{\frac{60.2}{36.1} -1}[/tex] = 3 Therefore λ = v/f = 343/36.1 = 9.5 m
L = 3*λ/4 = 3*9.5/4 = 7.12 m
The well is 7.12 m deep
