nless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. T

elchatosarapage

elchatosarapage

Answered question

2021-11-06

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. The fundamental frequency of a pipe that is open at both ends is 524 Hz. (1) How long is this pipe? If one end is now closed, find (2) the wavelength and (3) the frequency of the new fundamental.

Answer & Explanation

Twereen

Twereen

Beginner2021-11-07Added 13 answers

Step 1 
standing wave frequency in an exposed pipe:
f=nv2L 
The frequency of nth harmonic,  vThe velocity of the wave, 
n  nth harmonic  n=1,2,3,),L  length of the pipe 
Step 2 
1) 
fundamental freqencyfirst harmonic  n=1, 
f=nv2L 
L=nv2f1=1×344ms2×524Hz=0.328m 
Step 3 
standing wave frequency in a stopped pipe:
(a pipe which has a closed end) 
f=nv4L 
The frequency of nth harmonic,  vThe velocity of the wave, 
n  nth harmonic  n=1,2,3,),L  length of the pipe 
Step4 
2) 
fundamental freqencyfirst harmonic  n=1 
L=0.328m 
f=vλ=nv4L 
λ=4Ln=4×0.328m1=1.312m 
Step 5 
3) 
fundamental freqencyfirst harmonic  n=1 
L=0.328 
f=nv4L=1×344ms4×0.328m=262.2Hz 
Answer: 
1) 0.328m 
2) 1.312m 
3) 262.2HZ

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