Question

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

Answer 1

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = $\frac{2L}{3}$

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 = $0.25 x \frac{2L}{3}$

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

Distance = $0.25 x \frac{2 \times 39}{3}$

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.