Answer:
Part a)
55.3 cm
Part b)
41.5 cm
Explanation:
Pipe A is open at both ends so the fundamental frequency of this pipe is given as
[tex]f_o = \frac{V}{2L}[/tex]
here we know that
V = 343 m/s
[tex]f_o = 310 Hz[/tex]
now we have
[tex]310 = \frac{343}{2L}[/tex]
[tex]L = 55.3 cm[/tex]
Now we also know that second harmonic of pipe A and third harmonic of pipe B has same frequency
so we will have
[tex]\frac{2V}{2L_a} = \frac{3V}{4L_b}[/tex]
[tex]L_b = \frac{3}{4}L_a[/tex]
[tex]L_b = \frac{3}{4}(55.3) = 41.5 cm[/tex]