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Question

Physics Question on Waves

A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512Hz512\, Hz produces first resonance when the tube is filled with water to a mark 11cm11\, cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256Hz256\, Hz which produces first resonance when water reaches a mark 27cm27\, cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to:

A

328ms1328\,ms^{-1}

B

322ms1322\,ms^{-1}

C

341ms1341\,ms^{-1}

D

335ms1335\,ms^{-1}

Answer

328ms1328\,ms^{-1}

Explanation

Solution

λ14=11cm\frac{\lambda_{1}}{4}=11\, cm so ,
v512×4=11cm\frac{ v }{512 \times 4}=11 \,cm
λ24=27cm\frac{\lambda_{2}}{4}=27 \,cm so ,
v256×4=27cm\frac{ v }{256 \times 4}=27 \,cm
(2)(1)(2)-(1) v256×4×0.5=0.16\frac{ v }{256 \times 4} \times 0.5=0.16
v=0.16×2×4×256v =0.16 \times 2 \times 4 \times 256
v=328m/sv =328\, m / s
v256×4×0.5=0.16\frac{ v }{256 \times 4} \times 0.5=0.16
v=0.16×2×4×256v =0.16 \times 2 \times 4 \times 256
v=328m/sv =328 \,m / s