Question
Two open tubes, whose length are 50 cm and 50.05 cm respectively. 10 beats are produced in 3 seconds. Find their fundamental frequencies.
✓
Answer
Frequency of open tube
$\begin{aligned} n & =\frac{ v }{2 l} \\ v & \rightarrow \text { speed of sound } \\ l & \rightarrow \text { length of tube } \\ n_1 & =\frac{ v }{2 l_1}\ldots\ldots (1) \\ n_2 & =\frac{ v }{2 l_2}\ldots\ldots (2)\end{aligned}$
Given:
$\begin{aligned}n_1-n_2 & =\frac{10}{3} \\l_1 & =50 cm \\l_2 & =50.5 cm \\n_1-n_2 & =\frac{v}{2 l_1} \times \frac{v}{2 l_2} \\& =\frac{v}{2}\left[\frac{l_2-l_1}{l_1 l_2}\right] \\ \frac{10}{3} & =\frac{v}{2}\left[\frac{50.5-50}{50 \times 50.5}\right] \\\frac{10}{3} & =\frac{v}{2}\left[\frac{0.5}{2525}\right] \\\text { or } \quad v & =\frac{50500}{1.5}\end{aligned}$
$\begin{array}{l}=336.66 cm / s \\=336.66 m / \text { second }
\end{array}$
Put value in eqn. (1) and (2)
$\begin{array}{l}n_1=\frac{v}{2 l_1}=\frac{336.66}{2 \times 0.50} \\n_1=336.66 \cong 337 \text { vibration/sec. }\end{array}$
Similarly,
$\begin{array}{l}n_2=\frac{v}{2 l_2}=\frac{336.66}{2 \times 50.5 \times 10^{-2}} \\n_2=333 \text { vibration } / \text { second. }\end{array}$
$\begin{aligned} n & =\frac{ v }{2 l} \\ v & \rightarrow \text { speed of sound } \\ l & \rightarrow \text { length of tube } \\ n_1 & =\frac{ v }{2 l_1}\ldots\ldots (1) \\ n_2 & =\frac{ v }{2 l_2}\ldots\ldots (2)\end{aligned}$
Given:
$\begin{aligned}n_1-n_2 & =\frac{10}{3} \\l_1 & =50 cm \\l_2 & =50.5 cm \\n_1-n_2 & =\frac{v}{2 l_1} \times \frac{v}{2 l_2} \\& =\frac{v}{2}\left[\frac{l_2-l_1}{l_1 l_2}\right] \\ \frac{10}{3} & =\frac{v}{2}\left[\frac{50.5-50}{50 \times 50.5}\right] \\\frac{10}{3} & =\frac{v}{2}\left[\frac{0.5}{2525}\right] \\\text { or } \quad v & =\frac{50500}{1.5}\end{aligned}$
$\begin{array}{l}=336.66 cm / s \\=336.66 m / \text { second }
\end{array}$
Put value in eqn. (1) and (2)
$\begin{array}{l}n_1=\frac{v}{2 l_1}=\frac{336.66}{2 \times 0.50} \\n_1=336.66 \cong 337 \text { vibration/sec. }\end{array}$
Similarly,
$\begin{array}{l}n_2=\frac{v}{2 l_2}=\frac{336.66}{2 \times 50.5 \times 10^{-2}} \\n_2=333 \text { vibration } / \text { second. }\end{array}$
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