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

Question

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

Tardigrade Admin · Accepted Answer

When the dosed pipe is cut one open and one closed pipe are formed.
When pipe is closed at one end

n = \frac{v}{4 l}

Given,

n = 412 Hz

412 = \frac{v}{4 l} ... (i)

When pipe is cut into two equal halves then length of each is

\frac{l}{2}

.

n_{1} = \frac{v}{4 l}

(closed pipe)

n_{2} = \frac{v}{2 l}

(open pipe)
Where

l = \frac{l}{2}

∴ n_{1} = \frac{v}{4 ( \frac{l}{2} )}

Putting

v = 1648 l

from Eq. (i), we get

n_{1} = \frac{1648 l}{2 l} = 824 Hz

n_{2} = \frac{v}{2 ( \frac{l}{2} )} = \frac{v}{l} = \frac{1648 l}{l} = 1648 Hz

Note: A closed pipe produces only odd harmonics while an open pipe produces both even and odd harmonics.

Q. 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

$824 Hz, 1648 Hz$

$412 Hz, 824 Hz$

$206 Hz .412 Hz$

$216 Hz, 824 Hz$

Q. A pipe closed at one end produces a fundamental note of 412Hz. It is cut into two pieces of equal length the fundamental notes produced by the two pieces are

824Hz,1648Hz

412Hz,824Hz

206Hz.412Hz

216Hz,824Hz

Q. 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

$824 Hz, 1648 Hz$

$412 Hz, 824 Hz$

$206 Hz .412 Hz$

$216 Hz, 824 Hz$