Optimizing Resonance Length for Closed-End Pipe with a Tuning Fork

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SUMMARY

The shortest length of a closed-end pipe that resonates with a tuning fork of frequency 512 Hz is calculated to be 0.16 meters. This is derived using the formula L = 1/4λ, where λ is the wavelength. Given the speed of sound at 340 m/s, the relationship f = v/(4L) confirms the resonance length. The solution is accurate, with a minor rounding error noted in the discussion.

PREREQUISITES
  • Understanding of wave properties, specifically wavelength and frequency
  • Familiarity with the speed of sound in air (340 m/s)
  • Knowledge of resonance in closed-end pipes
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the relationship between frequency and wavelength in wave mechanics
  • Explore the concept of resonance in different types of pipes
  • Learn about the effects of temperature and pressure on the speed of sound
  • Investigate tuning fork applications in acoustics and musical instruments
USEFUL FOR

Physics students, acoustics researchers, and educators looking to deepen their understanding of sound resonance in closed-end pipes.

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Homework Statement


A tube is closed at one end by a piston which is slowly withdrawn as a tuning fork of frequency 512 Hz is sounded over it. What is the shortest length of pipe which will resonate with the fork?
(Speed of sound is 340m/s)


Homework Equations


L=1/4λ
f=v/(4L)

The Attempt at a Solution


Since the question wants the shortest length for resonance, so set L=1/4λ → 4L=λ
Then, f=v/(4L) → 512=340/(4L) → L=0.16m

Is the answer and the way i approached this question correct?
 
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Exhausted said:
)

The Attempt at a Solution


Since the question wants the shortest length for resonance, so set L=1/4λ → 4L=λ
Then, f=v/(4L) → 512=340/(4L) → L=0.16m

Is the answer and the way i approached this question correct?

It is correct but a little rounding error.

ehild
 
Thanks :D
 

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