A pipe closed at one end produces a fundamental note of 412 Hz. It is cut into two pieces of equal length,…
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 piece are
(a) 824 Hz, 1648 Hz
(b) 412 Hz, 824 Hz
(c) 206 Hz, 412 Hz
(d) 206 Hz, 824 Hz
Solution
Fundamental frequency, $f = \frac{v}{4l} = 412\text{ Hz}$
[Diagram shows a closed pipe cut into two equal halves, resulting in an open pipe producing frequency $f_1$ and a closed pipe producing frequency $f_2$.]
After cutting, frequencies of fundamental notes produced,
$f_1 = \frac{v}{2(l / 2)} = \frac{v}{l} = 4 \times 412 = 1648\text{ Hz}$
and $f_2 = \frac{v}{4(l / 2)} = \frac{v}{2l} = 2 \times 412 = 824\text{ Hz}$