Question

UUTUU
(ii) A tube of certain diameter and of length 48 cm is open at
both ends. Its fundamental frequency of resonance is found
to be 320 Hz. The velocity of sound in air is 320 m/s.
Estimate the diameter of the tube. One end of the tube is
now closed. Calculate the lowest frequency of resonance for
the tube.
[fundesmental faquency of flored​

Answer 1

Answer:

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Explanation:

Answer 2

ANSWER

λ=fv=320320=1m

As it is open at both ends correction at each end is 0.6r.

∴2λ 0.6r−+ 0.6r=l

50 −+ 1.2r=48 cm

1.2r=2cm

r=1.22=1.65

∴d=2r=3.33 cm

Thanks...