Math Problem Statement

When two closed tubes (closed at one end) of different lengths each resonate at their 3rd harmonic, a 15.9 Hz beat frequency is observed. Both tubes are now opened at both ends. When each tube now resonates at its n = 3 harmonic frequency, the observed beat frequency is ______ Hz.

Solution

To solve this problem, we'll break it down step by step.

Step 1: Understand the Situation for Closed Tubes

When tubes are closed at one end, they resonate at odd harmonics (1st, 3rd, 5th, etc.). The frequency for the nn-th harmonic of a closed tube is given by:

fn=nv4Lf_n = \frac{n \cdot v}{4L}

where:

  • nn is the harmonic number,
  • vv is the speed of sound in air,
  • LL is the length of the tube.

Given that both tubes resonate at their 3rd harmonic, the frequencies for the two tubes can be written as:

f1=3v4L1f_1 = \frac{3 \cdot v}{4L_1} f2=3v4L2f_2 = \frac{3 \cdot v}{4L_2}

The beat frequency between these two frequencies is given by:

Beat frequency=f1f2=15.9 Hz\text{Beat frequency} = |f_1 - f_2| = 15.9 \text{ Hz}

Step 2: Understand the Situation for Open Tubes

When both ends of the tubes are open, they resonate at all harmonics (1st, 2nd, 3rd, etc.). The frequency for the nn-th harmonic of an open tube is given by:

fn=nv2Lf_n = \frac{n \cdot v}{2L}

For the n=3n = 3 harmonic:

f1=3v2L1f_1' = \frac{3 \cdot v}{2L_1} f2=3v2L2f_2' = \frac{3 \cdot v}{2L_2}

Step 3: Compare the Beat Frequencies

The ratio of the beat frequencies for the open and closed tubes can be determined by the ratio of the corresponding frequency expressions:

Beat frequency (open)Beat frequency (closed)=3v2L13v2L23v4L13v4L2=2\frac{\text{Beat frequency (open)}}{\text{Beat frequency (closed)}} = \frac{\frac{3v}{2L_1} - \frac{3v}{2L_2}}{\frac{3v}{4L_1} - \frac{3v}{4L_2}} = 2

This means that the beat frequency when the tubes are open at both ends will be twice the beat frequency when they were closed at one end.

Step 4: Calculate the Beat Frequency for Open Tubes

Given that the beat frequency for closed tubes is 15.9 Hz, the beat frequency for open tubes will be:

Beat frequency (open)=2×15.9 Hz=31.8 Hz\text{Beat frequency (open)} = 2 \times 15.9 \text{ Hz} = 31.8 \text{ Hz}

Final Answer:

The observed beat frequency when each tube resonates at its n=3n = 3 harmonic frequency with both ends open is 31.8 Hz.


Do you have any questions, or would you like further details on any step? Here are some related questions you might explore:

  1. What is the difference between harmonics in closed and open tubes?
  2. How does the length of a tube affect the frequency of its resonant harmonics?
  3. What is the significance of the beat frequency in sound waves?
  4. How would the beat frequency change if the tubes were of equal length?
  5. What is the formula for beat frequency, and how is it derived?
  6. How does the speed of sound in air affect resonant frequencies in tubes?
  7. What happens to the harmonics if the tubes are filled with a different gas?
  8. How would the beat frequency be affected if one tube were closed and the other open?

Tip: Always check the boundary conditions of a problem (like open vs. closed tubes) as they can significantly alter the equations and outcomes.

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Math Problem Analysis

Mathematical Concepts

Acoustics
Resonance
Harmonics

Formulas

Closed tube harmonic frequency: fn = (n * v) / (4 * L)
Open tube harmonic frequency: fn = (n * v) / (2 * L)

Theorems

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Suitable Grade Level

Advanced High School