The harmonic series from a long tube is given below. Isthis tube acting as an open-pipe resonator or a closed-piperesonator? Explain your answer. 203 Hz, 609 Hz, 1015 Hz, 1421 Hz

L=1.7604

Explanation:

Three successive harmonics in the pipe is given as

f_1 = 435 Hzf1=435Hz

f_2 = 725 Hzf2=725Hz

f_3 = 1015 Hzf3=1015Hz

now if we take the ratio of all frequency

then we have

f1 : f2 : f3 = 435 : 725 : 1015 = 3 : 5 : 7

since the ratio of consecutive frequency is in ratio of odd

numbers so this must be 3rd harmonics, 5th

harmonics and 7 harmonics

And as per the formula we have

f_1 = / frac{3v}{4L}f1=4L3v

435 = / frac{3 (340) }{4L}435=4L3 (340)

L = 1.706 mL=1.706m