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A tuning fork with frequency $800 \mathrm{~Hz}$ produces resonance in a resonance column tube with upper end open and lower end closed by water surface. Successive resonance are observed at length $9.75 \mathrm{~cm}, 31.25 \mathrm{~cm}$ and $52.75 \mathrm{~cm}$. The speed of sound in air is
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The correct answer is:
$344 \mathrm{~m} / \mathrm{s}$
For vibrating tunning fork over a resonance tube, the first resonance is obtained at the length
$$
l_1=\frac{\lambda}{4}
$$
and for second resonance,
$$
\mathrm{l}_2=\frac{\lambda}{4}+\frac{\lambda}{2}=\frac{3 \lambda}{4}
$$
From Eq. (i) and (ii), we get
$$
\begin{array}{cc}
& \mathrm{l}_2-\mathrm{l}_1=\frac{3 \lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2} \\
\Rightarrow & \lambda=2\left(l_2-l_1\right) \\
\Rightarrow & \mathrm{v}=2 \mathrm{f}\left(\mathrm{l}_2-\mathrm{l}_1\right)
\end{array}
$$
..(iii) $\left(\because \lambda=\frac{\mathrm{v}}{\mathrm{f}}\right)$
Here, $\mathrm{f}=800 \mathrm{~Hz}, \mathrm{l}_1=9.75 \mathrm{~cm}, \mathrm{l}_2=31.25 \mathrm{~cm}$
Substituting the given values in Eq. (iii), we get
$$
\begin{aligned}
& \Rightarrow \mathrm{v}=2 \times 800(31.25-9.75) \\
& =34400 \mathrm{~cm} / \mathrm{s}=344 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
$$
l_1=\frac{\lambda}{4}
$$
and for second resonance,
$$
\mathrm{l}_2=\frac{\lambda}{4}+\frac{\lambda}{2}=\frac{3 \lambda}{4}
$$
From Eq. (i) and (ii), we get
$$
\begin{array}{cc}
& \mathrm{l}_2-\mathrm{l}_1=\frac{3 \lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2} \\
\Rightarrow & \lambda=2\left(l_2-l_1\right) \\
\Rightarrow & \mathrm{v}=2 \mathrm{f}\left(\mathrm{l}_2-\mathrm{l}_1\right)
\end{array}
$$
..(iii) $\left(\because \lambda=\frac{\mathrm{v}}{\mathrm{f}}\right)$
Here, $\mathrm{f}=800 \mathrm{~Hz}, \mathrm{l}_1=9.75 \mathrm{~cm}, \mathrm{l}_2=31.25 \mathrm{~cm}$
Substituting the given values in Eq. (iii), we get
$$
\begin{aligned}
& \Rightarrow \mathrm{v}=2 \times 800(31.25-9.75) \\
& =34400 \mathrm{~cm} / \mathrm{s}=344 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
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