A closed organ pipe and an open organ pipe of same length produce 2 beats per second when they are set into…
A closed organ pipe and an open organ pipe of same length produce 2 beats per second when they are set into vibrations together in fundamental mode. The length of open pipe is now halved and that of closed pipe is doubled. The number of beats produced per second will be
4
3
8
7
Solution
Fundamental frequency of closed pipe, $\mathrm{n}_{\mathrm{c}}=\frac{\mathrm{V}}{4 \mathrm{~L}}$ Fundamental frequency of open pipe, $\mathrm{n}_0=\frac{\mathrm{v}}{2 \mathrm{~L}}$ They produce 2 beats per second
$\begin{aligned} & \therefore \mathrm{n}_{\mathrm{o}}-\mathrm{n}_{\mathrm{c}}=2, \\ & \frac{\mathrm{V}}{2 \mathrm{~L}}-\frac{\mathrm{V}}{4 \mathrm{~L}}=2 \\ & \frac{\mathrm{V}}{4 \mathrm{~L}}=2 \text { or } \frac{\mathrm{v}}{\mathrm{L}}=8\end{aligned}$
When length of open pipe is halved
$\mathrm{n}_{\mathrm{o}}^{\prime}=\frac{\mathrm{v}}{2\left(\frac{\mathrm{L}}{2}\right)}=\frac{\mathrm{v}}{\mathrm{L}}$
When length of closed pipe is doubled
$\mathrm{n}_{\mathrm{c}}^{\prime}=\frac{\mathrm{v}}{4 \times 2 \mathrm{~L}}=\frac{\mathrm{v}}{8 \mathrm{~L}}$
New beat frequency $=\mathrm{n}_{\mathrm{o}}^{\prime}-\mathrm{n}_{\mathrm{c}}^{\prime}=\frac{\mathrm{v}}{\mathrm{L}}-\frac{\mathrm{v}}{8 \mathrm{~L}}=\frac{7 \mathrm{v}}{8 \mathrm{~L}}=\frac{7}{8} \times 8=7$
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