A tube that is open at both ends supports a standing wave with harmonics at 300 Hz and 400 Hz, with no harmonics between.

A tube that is open at both ends supports a standing wave with harmonics at 300 Hz and 400 Hz, with no harmonics between. What is the fundamental frequency of this tube?
A. 50 Hz  
B. 100 Hz
C. 150 Hz
D. 200 Hz
E. 300 Hz

The Correct answer and Explanation is:

To determine the fundamental frequency of a tube that is open at both ends and supports harmonics at 300 Hz and 400 Hz with no harmonics in between, we need to understand the relationship between harmonics and the fundamental frequency in open tubes.

Explanation:

1. Fundamental Frequency and Harmonics: In a tube open at both ends, the harmonics are integer multiples of the fundamental frequency (f₁). The frequencies of the harmonics can be expressed as:
   [
   f_n = n \cdot f_1
   ]
   where (n) is a positive integer (1 for the fundamental, 2 for the first harmonic, and so on).
2. Given Frequencies: The frequencies provided are 300 Hz and 400 Hz. To find the fundamental frequency, we need to analyze how these frequencies relate to the fundamental.
3. Finding the Difference: The difference between the two given frequencies is:
   [
   400 \text{ Hz} – 300 \text{ Hz} = 100 \text{ Hz}
   ]
   This indicates that the two harmonics represent consecutive harmonics of the fundamental frequency.
4. Finding the Harmonic Number: Since there are no harmonics between 300 Hz and 400 Hz, they must correspond to consecutive harmonic numbers. If we assume (f_1) is the fundamental frequency, then:

• (300 \text{ Hz} = n \cdot f_1)
• (400 \text{ Hz} = (n+1) \cdot f_1)

1. Setting up Equations:
   [
   n \cdot f_1 = 300 \text{ Hz}
   ]
   [
   (n + 1) \cdot f_1 = 400 \text{ Hz}
   ] Rearranging the second equation gives:
   [
   n \cdot f_1 + f_1 = 400 \text{ Hz}
   ]
2. Solving for (f_1): Substituting (n \cdot f_1 = 300) into the second equation:
   [
   300 + f_1 = 400
   ]
   [
   f_1 = 400 – 300 = 100 \text{ Hz}
   ]

Thus, the fundamental frequency of the tube is 100 Hz.

Conclusion:

The answer is B. 100 Hz. This represents the lowest frequency that can resonate within the tube, and it supports the harmonics at 300 Hz and 400 Hz, which are the second and third harmonics, respectively.