Answer:
The frequency of its second harmonic is 475.14 Hz.
Explanation:
Given;
length of the open tube, L = 72.4 cm = 0.724 m
speed of sound, v = 344 m/s
the wavelength at fundamental frequency is calculated as;
L = A -----N + N----A
where;
L is length of the tube
A is antinode of wave in the pipe
N is node of the wave in the pipe
[tex]L = \frac{\lambda}{4} + \frac{\lambda}{4}\\\\L = \frac{\lambda}{2}\\\\\lambda = 2L = \lambda_o[/tex]
the fundamental frequency of the wave is calculated as;
[tex]f = \frac{v}{\lambda}\\\\f_o =\frac{v}{\lambda _o} \\\\f_o = \frac{v}{2L} \\\\f_o = \frac{344}{2\times 0.724}\\\\f_o = 237.57 \ Hz[/tex]
the frequency of the second harmonic is calculated as;
2f₀ = 2 x 237.57
2f₀ = 475.14 Hz.
