A tube \({1.0}{m}\) long is closed at one end. A stretched wire is placed near the open end. The wire is \({0.3}{m}\) long and a mass of \({0.01}{k}{g}\) . It is held fixed at both ends and vibrates in its fundamental mode. It sets the air column in the tube into vibration at its fundamental frequency by resonance. Find

(a) the frequency of oscillation of the air column and

(b) the tension in the wire.

Speed of sound in air = \({330}{m}/{s}\) .

(a) the frequency of oscillation of the air column and

(b) the tension in the wire.

Speed of sound in air = \({330}{m}/{s}\) .

Answer: A、B

(a) fundamental frequency of closed pipe = \(\frac{{\upsilon}}{{{4}{l}}}\)

= \(\frac{{{330}}}{{{4}\times{1}}}={82.5}{H}_{{{Z}}}\)

At resonance, give that :

fundamental frequency of stretched wire (at both ends)

= fundamental frequency of air column

\(\therefore\frac{{\upsilon}}{{{2}{l}}}={82.5}{H}_{{{Z}}}\)

\(\therefore\frac{\sqrt{{{\left({T}/\mu\right)}}}}{{{2}{l}}}={82.5}{H}_{{{Z}}}\)

or \({T}=\mu{\left({2}\times{0.3}\times{82.5}\right)}^{{{2}}}\)

= \({\left(\frac{{{0.01}}}{{{0.3}}}\right)}{\left({2}\times{0.3}\times{82.67}{.5}\right)}^{{{2}}}\)

= \({81.675}{N}\)

(a) fundamental frequency of closed pipe = \(\frac{{\upsilon}}{{{4}{l}}}\)

= \(\frac{{{330}}}{{{4}\times{1}}}={82.5}{H}_{{{Z}}}\)

At resonance, give that :

fundamental frequency of stretched wire (at both ends)

= fundamental frequency of air column

\(\therefore\frac{{\upsilon}}{{{2}{l}}}={82.5}{H}_{{{Z}}}\)

\(\therefore\frac{\sqrt{{{\left({T}/\mu\right)}}}}{{{2}{l}}}={82.5}{H}_{{{Z}}}\)

or \({T}=\mu{\left({2}\times{0.3}\times{82.5}\right)}^{{{2}}}\)

= \({\left(\frac{{{0.01}}}{{{0.3}}}\right)}{\left({2}\times{0.3}\times{82.67}{.5}\right)}^{{{2}}}\)

= \({81.675}{N}\)

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