Answer
See answers.
Work Step by Step
We use the effective length of the tube.
$$\ell_{eff}=\ell+\frac{1}{3}D=0.55m+\frac{1}{3}0.030m=0.56m$$
Use the equation for the fundamental frequency of a tube closed at one end,. See figure 12-12.
First harmonic, first uncorrected, then taking the end correction into consideration
$$f=\frac{v}{4L}=\frac{343m/s}{4(0.55m)}=156Hz$$
$$f=\frac{v}{4L}=\frac{343m/s}{4(0.56m)}=153Hz\approx 150Hz$$
Second harmonic, first uncorrected, then taking the end correction into consideration
$$f=\frac{3v}{4L}=\frac{3(343m/s)}{4(0.55m)}=468Hz$$
$$f=\frac{3v}{4L}=\frac{3(343m/s)}{4(0.56m)}=459Hz\approx 460Hz$$
Third harmonic, first uncorrected, then taking the end correction into consideration
$$f=\frac{5v}{4L}=\frac{5(343m/s)}{4(0.55m)}=780Hz$$
$$f=\frac{5v}{4L}=\frac{5(343m/s)}{4(0.56m)}=766Hz\approx 770Hz$$
Fourth harmonic, first uncorrected, then taking the end correction into consideration
$$f=\frac{7v}{4L}=\frac{7(343m/s)}{4(0.55m)}=1091Hz$$
$$f=\frac{7v}{4L}=\frac{7(343m/s)}{4(0.56m)}=1072Hz\approx 1100Hz$$
