When the air column of a resonance tube is vibrated together with a tuning fork, 3 beats are heard per…

When the air column of a resonance tube is vibrated together with a tuning fork, 3 beats are heard per second, either the temperature of the air column is \(51^{\circ} \mathrm{C}\) or \(16^{\circ} \mathrm{C}\). The frequency of the tuning fork is
  1. \(128 \mathrm{~Hz}\)
  2. \(98 \mathrm{~Hz}\)
  3. \(105 \mathrm{~Hz}\)
  4. \(256 \mathrm{~Hz}\)

Solution

Number of beats per second when the air column of resonance tube, \(n=3\) If \(n\) be the frequency of tuning fork, Then, \(n \propto \sqrt{T}[T \rightarrow\) temperature \(]\) \(\frac{n_1}{n_2}=\sqrt{\frac{T_1}{T_2}}\) At \(51^{\circ} \mathrm{C}, T_1=273+51=324 \mathrm{~K}\) \(n_1=n+3\{\) frequency of tuning fork increase at the higher temperature \} At \(16^{\circ} \mathrm{C}, T_2=273+16=289 \mathrm{~K}\) \(n_2=n-3\) [At lower temperature, frequency of the tuning fork decreases.] \(\therefore\) From Eq. (i), \(\begin{array}{ll} & \frac{n_1}{n_2}=\sqrt{\frac{324}{289}} \Rightarrow \frac{n+3}{n-3}=\frac{18}{17} \\ & 18 n-54=17 n+51 \\ \Rightarrow \quad 18 n-17 n=54+51 \Rightarrow n=105 \mathrm{~Hz} \end{array}\)

Asked in: AP EAMCET 2019 (22 Apr Shift 1)

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