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14.Waves and Sound
medium
A pipe closed at one end produces a fundamental note of $412\,Hz.$ It is cut into two pieces of equal length the fundamental notes produced by the two pieces are
A
$824\,Hz,\,\,1648\,Hz$
B
$412\,Hz,\,\,824\,Hz$
C
$206\,Hz,\,\,412\,Hz$
D
$216\,Hz,\,\, 824\,Hz$
Solution
Given $\frac{v}{4 L}=412$
$\frac{\mathrm{v}}{\mathrm{L}}=1648$
When pipe is cut we will get one $COP$ and one $\mathrm{OOP}$
$\therefore $ Fundamental frequency of $COP$
$=\frac{\mathrm{v}}{4 \mathrm{L}}=\frac{\mathrm{v}}{4\left(\frac{\mathrm{L}}{2}\right)}=\frac{\mathrm{v}}{2 \mathrm{L}}=\frac{1648}{2}=824 \mathrm{\,Hz}$
And fundamental frequency of $OOP$
$=\frac{\mathrm{v}}{2 \mathrm{L}^{\prime}}=\frac{\mathrm{v}}{2\left(\frac{\mathrm{L}}{2}\right)}=\frac{\mathrm{v}}{\mathrm{L}}=1648 \mathrm{\,Hz}$
Standard 11
Physics
