Gujarati
Hindi
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

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