Math Problem Statement

Mech_and_Wave_Exercise_11.pdf

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solve 3

Solution

The file contains several physics problems and their corresponding answers. Here's a detailed breakdown of the requested problem and solution:

Problem 3

A primitive flute is designed with a length of 40 cm and has two finger holes at 10 cm and 20 cm from the mouthpiece. The task is to calculate and draw the lowest frequencies when:

  • a) Both finger holes are covered.
  • b) The 10 cm finger hole is covered.
  • c) The 20 cm finger hole is covered.
  • d) Both finger holes are open.

Given Data:

  • Flute length: L=40cmL = 40 \, \text{cm} or 0.4m0.4 \, \text{m}
  • Speed of sound in air: v=340m/sv = 340 \, \text{m/s}

Solution:

  1. General Formula for Frequency:

    • For a tube with one open end: f=nv4Lf = \frac{nv}{4L}, where n=1,3,5,n = 1, 3, 5, \dots (odd harmonics)
    • For a tube with two open ends: f=nv2Lf = \frac{nv}{2L}, where n=1,2,3,n = 1, 2, 3, \dots (all harmonics)
  2. a) Both Finger Holes Covered:

    • This is equivalent to a closed pipe of length L=0.4mL = 0.4 \, \text{m}.
    • The fundamental frequency is: f1=v4L=3404×0.4=425Hz.f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.4} = 425 \, \text{Hz}.
  3. b) 10 cm Finger Hole Covered:

    • Effective length becomes L=0.3mL = 0.3 \, \text{m} (40 cm – 10 cm).
    • The fundamental frequency is: f1=v4L=3404×0.3=850Hz.f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.3} = 850 \, \text{Hz}.
  4. c) 20 cm Finger Hole Covered:

    • Effective length becomes L=0.2mL = 0.2 \, \text{m} (40 cm – 20 cm).
    • The fundamental frequency is: f1=v4L=3404×0.2=1700Hz.f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.2} = 1700 \, \text{Hz}.
  5. d) Both Finger Holes Open:

    • The flute behaves as an open pipe with L=0.4mL = 0.4 \, \text{m}.
    • The fundamental frequency is: f1=v2L=3402×0.4=425Hz.f_1 = \frac{v}{2L} = \frac{340}{2 \times 0.4} = 425 \, \text{Hz}.

Final Results:

  • a) 425 Hz
  • b) 850 Hz
  • c) 1700 Hz
  • d) 1700 Hz

If you'd like additional explanation or diagrams for the standing waves, let me know!


Expansion Questions:

  1. How do the harmonics of an open pipe compare to those of a closed pipe?
  2. What happens to the frequency if the speed of sound changes (e.g., different temperatures)?
  3. How would damping in the flute affect the observed frequencies?
  4. Can you derive the formula for frequency from basic wave principles?
  5. How does the presence of finger holes impact the flute’s resonance and sound quality?

Tip:

For standing wave problems, always visualize the placement of nodes and antinodes to understand the effective length changes!

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Math Problem Analysis

Mathematical Concepts

Wave Motion
Acoustics
Standing Waves

Formulas

f = v / 4L for tubes with one open end
f = v / 2L for tubes with two open ends

Theorems

Harmonic Frequency Relationships in Resonance Tubes

Suitable Grade Level

Undergraduate Physics