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Helly123
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Homework Statement
That is a speaker. Sound wave is sent out from spesker S into pipe of uniform thickness.
Piston P move to left
1st resonance at 0.045 m
2nd resonance at 0.151 m
Frequency of the sounx 1620 Hz
Piston is stopped at the position that 2nd resonance occur
Then, frequency increased litle by litle
Find frequency at which resonance occur again
End correction is 0.008 m
##\lambda## = 0.21 m
Velocity = 340 m/s
Homework Equations
The Attempt at a Solution
Frequency = Velocity/##\lambda##
L2 = 0.151 m = 3/4 ##\lambda##
##\lambda## = 4/3 (0.151)m =0.2
Frequency = 340 / 0.20 = 1700 Hz
How to find frequecy when resonance occur again?
What will change the frequency, the velocity, or ##\lambda## ?
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