An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400hz. what is the distance from the center of the pipe to the nearest antinode. a. 13cm b. 9.8cm c. 6.5cm d. 3.2cm e. 0cm

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength =

The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength

Distance from center  = 0.25 × wavelength

Distance =

Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm

Distance =

Distance from the center to the nearest anti - node = 6.5 cm

Hence, the nearest distance to the anti - node from the center = 6.5 cm

So, option C is correct.

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explanation: