Superpositon and Standing Waves problem

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SUMMARY

The discussion centers on determining sound frequencies at which a child with a 1.3 cm ear canal experiences increased hearing sensitivity, specifically using the formula for an open-closed tube, f = m(v/4L). The user initially calculated frequencies of 6600 Hz and 20000 Hz using m values of 1 and 3, respectively. However, they sought clarification on selecting appropriate nodes (m) for accurate frequency determination. The correct approach confirms that the next resonance frequency is indeed three times the first, aligning with the audible range of 20,000 Hz.

PREREQUISITES
  • Understanding of wave mechanics, specifically sound waves.
  • Familiarity with the properties of open-closed tubes in acoustics.
  • Knowledge of the formula f = m(v/4L) for calculating frequencies.
  • Awareness of the audible frequency range (20 Hz to 20,000 Hz).
NEXT STEPS
  • Research the properties of open-closed tubes and their resonance frequencies.
  • Study the relationship between wavelength and frequency in sound waves.
  • Explore the concept of nodes and antinodes in wave mechanics.
  • Investigate the effects of ear canal length on hearing sensitivity.
USEFUL FOR

Students studying acoustics, audio engineers, and audiologists interested in the relationship between ear canal dimensions and sound frequency sensitivity.

weezer13578
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Homework Statement


A child has an ear canal that is 1.3 cm long. At what sound frequencies in the audible range will the child have increased hearing sensitivity?


Homework Equations


for open-closed tube, f= m(v/4L)


The Attempt at a Solution


I need two frequencies since the question is asking for a rage in which the child has increased hearing sensitivity. So I thought for an open-closed tube, I could use f= m(v/4L) where m=1 and 3. But my answer does not work. How do I know what nodes (m) to choose? Please help.
 
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I have previously found 6600 and 2.0*10^4 using m=1 and 3, and this range didn't work.
Could anyone help?
 
The question is asking for specific frequecies in the audible range; not a range of frequencies.
The method is correct. 6600Hz is in the audible range (quarter wavelength). The next resonance should be at 3 times that value. (3/4 wavelength)
Audible range goes up to 20000Hz typically.
 

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