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# Physics Questions

Assume the speed of sound is 331 m/s.

1. You have an open tube that is 1.3 m long.
a. What is the fundamental frequency and wavelength of the tube?
b. What is the third harmonic frequency and wavelength?

2. You have a closed tube that is 1.3 m long.
a. What is the fundamental frequency and wavelength of the tube?
b. What is the third harmonic frequency and wavelength?

3. How will you know when your tube is in resonance?

### 1 Answer by Expert Tutors

Amarjeet K. | Professional Engineer for Math and Science TuroringProfessional Engineer for Math and Scien...
4.6 4.6 (8 lesson ratings) (8)
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Wave speed = frequecy x wavelength

Frequency = wave speed/wavelength

1. In an open tube antinodes are located at the ends of open tube. Minimum number of waves that can fit in an open tube with minimum number of nodes and antinodes is λ/2. Next higher number of wavelengths would be λ, 3/2λ, 2λ.......

λ/2 = 1.3m
λ= 2.6m
Fundamental frequency = 331/2.6 = 127.3 Hz

second harmonic = 331/1.6 =206.8Hz
Third harmonic = 331/(1.6/2) = 413.75Hz

2. In closed tube a node is present at closed end and antinode at open end.
Minimum number of waves that can fit in an open tube with minimum number of nodes and antinodes is λ/4.
Next harmonic will be with 3/4λ, 1.25λ, 1.75λ

For fundamental frequecy λ/4 = 1.6, λ=6.4m

for 3/4λ = 1.6, λ=2.13m
for 1.25λ = 1.6, λ=1.28m

Fundamental frequecy = 331/6.4 = 51.71 Hz
second harmonic  - none because next harmonic is three times the fundamental frequeny
third harmonic frequency = 331/2.13=155.4
fifth harmonic = 331/1.28=258.6Hz = five times the fundamental frequency.

3. If you hold an activated tuning fork9 or any other source of sound waves) in top of a resonanting tube. You would hear a large sound when the adjusting air column will resonate with the frequency of the tuning fork.

:)