Frequency for resonance to occur

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The discussion revolves around calculating the frequency at which resonance occurs again after the piston is stopped at the second resonance position in a pipe. Initial resonances were identified at 0.045 m and 0.151 m, with a frequency of 1620 Hz and an end correction of 0.008 m. The participants explored the relationship between frequency, wavelength, and the speed of sound, concluding that increasing frequency decreases wavelength. Calculations led to a proposed frequency of 1700 Hz for the third resonance, but discrepancies in values prompted discussions about potential round-off errors and assumptions in the calculations. Ultimately, the conversation highlighted the complexity of resonance in sound waves within a uniform pipe.
  • #31
Helly123 said:
Maybe
##L_3(\lambda')## = ##\frac{5}{4}\lambda## - ε
0.151 = ##\frac{5}{4}\lambda## - 0.008
##\frac{5}{4}\lambda## = 0.159
##\lambda## = 0.2 * 4/5 = 0.16 = 0.2 m

That way, frequency = 340/0.2 = 1700 Hz
Here is what I think happened. Whoever made the answer key thought ##L_3(λ') = \frac 4 5 λ'-ε##. That would give 1710hz for the frequency. Rounding to 2 significant digits, it would be 1700hz.
 
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  • #32
tnich said:
Here is what I think happened. Whoever made the answer key thought ##L_3(λ') = \frac 4 5 λ'-ε##. That would give 1710hz for the frequency. Rounding to 2 significant digits, it would be 1700hz.
How can you assume the 1710 Hz? How to get that result?
 
  • #33
Helly123 said:
How can you assume the 1710 Hz? How to get that result?
Using ##\frac 4 5 λ## instead of ##\frac 5 4 λ##, but that would not be correct.
 
  • #34
Ok thanks
 

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