A pipe of length 100 cm is open at both ends. A loudspeaker situated at one end of the pipe can emit sound of different wavelengths.

Question 21

A pipe of length 100 cm is open at both ends. A loudspeaker situated at one end of the pipe can emit sound of different wavelengths.

At which wavelength can a stationary wave be produced in the pipe?

A 50 cm                      B 75 cm                      C 150 cm                    D 300 cm

Reference: Past Exam Paper – June 2017 Paper 13 Q26

Solution:

Answer: A. 

The pipe is opened at both ends. So, antinodes would be located at the ends of the pipe.

The fundamental mode consist of an antinode at both ends (antinode-node-antinode (ANA)). In this case (if you draw the waves), half a wavelength occupies the length of the pipe.

λ / 2 = 100 cm

Wavelength λ = 2 × 100 = 200 cm

This is not an available choice.

The next mode (1^st overtone) would consist of antinodes at the ends and an antinode inside the tube (ANANA). In this case, one complete wave occupies the length of the tube.

λ = 100 cm

This is also not an available choice.

The next mode (2^nd overtone) would consist of antinodes at the ends and 2 antinodes inside the tube (ANANANA). In this case, one and a half wavelength occupies the length of the tube.

3 λ / 2 = 100

λ = 200 / 3 = 66.7 cm

This is also not an available choice.

The next mode (3^rd overtone) would consist of antinodes at the ends and 3 antinodes inside the tube (ANANANANA). In this case, one and a half wavelength occupies the length of the tube.

2 λ = 100

λ = 100 / 2 = 50 cm

Alternatively, we could consider a general formula for the wavelength in open-open pipes. (From your notes,) this would be

(for the (n-1)^th overtone)

λ = 2L / n

Trying values of n (= 1, 2, 3, 4, 5, …) would give the answer.