Calculating the speed of sound in air

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SUMMARY

The speed of sound in air can be calculated using resonance lengths observed in a closed pipe. In this discussion, a tuning fork with a frequency of 325 Hz was used, and the first two resonance lengths were measured at 25.4 cm and 77.4 cm. The effective length between the first two anti-nodes is 52 cm, leading to the calculation of speed using the formula v = f * (2 * Length). The correct speed of sound determined from these parameters is 338 m/s.

PREREQUISITES
  • Understanding of wave mechanics and resonance
  • Familiarity with the formula v = f * λ
  • Knowledge of closed pipe acoustics
  • Basic proficiency in unit conversions (cm to m)
NEXT STEPS
  • Study the principles of resonance in closed pipes
  • Learn about the relationship between frequency and wavelength in sound waves
  • Explore advanced calculations of sound speed in different mediums
  • Investigate the effects of temperature and pressure on the speed of sound
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Students studying physics, particularly those focusing on wave mechanics and acoustics, as well as educators looking for practical examples of sound speed calculations.

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Homework Statement


In a resonance tube, using a tuning fork of frequency 325 Hz,the first two resonance lengths are observed at 25.4cm and 77.4cm.The speed of sound in air is
1. 318m/s 2. 338m/s 3.358m/s 4.378m/s 5. 368m/s


Homework Equations



v=f*lambda

The Attempt at a Solution


Since the question states 25.4cm and 77.4cm as the 1st two resonance lengths,25.4cm has to be the length of the tube at the fundamental frequency and 77.4cm at the 1st overtone.
And since 77.4cm is approximately 3 times the fundamental length of 25.4cm,I'm assuming it's a closed pipe.

Therefore, lambda/4 = 0.254m
lambda=1.016m
Then using v=f * lambda
=325*1.016
I get v =330.2m/s which is not among the choices given, so I'm really hoping someone could help me out here.
Thanks
 
Last edited:
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Think about the question again;

"the first two resonance lengths are observed at 25.4cm & 77.4cm" means the distance between the first two anti-nodes = 52cm, so effectively the "length" in the equation is 52

frequency = [tex]\frac{Speed}{2 * Length}[/tex]
325 = [tex]\frac{Speed}{2 * 0.52}[/tex]
 
Thanks. I get it now. I get a velocity of 338m/s
 

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