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

AI Thread Summary
The discussion focuses on calculating the shortest length of a closed-end pipe that resonates with a tuning fork at a frequency of 512 Hz, using the speed of sound at 340 m/s. The formula L = 1/4λ is applied, leading to the equation f = v/(4L). The calculated length is determined to be 0.16 meters, which is confirmed to be correct with a minor rounding error noted. The approach to the problem is validated, ensuring the solution aligns with the physics principles involved. Overall, the method used effectively demonstrates the resonance characteristics of the closed-end pipe.
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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A tube is closed at one end by a piston which is slowly withdrawn as a tuning fork of frequency 512 Hz is sounded over it.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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