An organic pipe closed at one end is 80cm long. Determine the frequency of the fundamental note assuming that the speed of sound in air is 340ms-1
\(\lambda\) = 4L
= 4 x 80 = 320cm
Converting 320cm to m, we have 3.2m
F = \(\frac{V}{\lambda}\)
= \(\frac{340}{3.2}\)
= 106.25Hz
\(\approxeq\) 106Hz
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