Question

For a certain organ pipe, three successive resonant frequencies are observed at $425{\text{Hz}}$ , $595{\text{Hz}}$ and $765{\text{Hz}}$. The speed of sound in the air is $340{\text{m}}{{\text{s}}^{ - 1}}$ . The pipe is a
A) closed pipe of length 1 m.
B) closed pipe of length 2 m.
C) open pipe of length 1 m.
D) open pipe of length 2 m.

Answer 1

A closed pipe will generate odd harmonics while an open pipe will generate even harmonics. Taking the ratio of the given three successive resonant frequencies will help us to determine whether the organ pipe is a closed one or an open pipe. Once we figure out whether the pipe is closed or open, we can easily determine its length using the relation for the fundamental frequency of the pipe.

Formula used:
-The length of a closed pipe is given by, $l = \dfrac{v}{{4f}}$ where $v$ is the speed of sound and $f$ is the fundamental frequency.

Complete step by step solution:
Step 1: Express the ratio of the three successive resonant frequencies.
The three resonant frequencies are given to be $425{\text{Hz}}$ , $595{\text{Hz}}$ and $765{\text{Hz}}$ .
Their ratio can be represented as $425{\text{Hz : }}595{\text{Hz : 765Hz}}$ .
On simplifying the above ratio we obtain, $425:595:765 = 5:7:9$
It can also be expressed as $425:595:765 = 5:\left( {5 + 2} \right):\left( {5 + 4} \right)$
This suggests that the harmonics are odd harmonics where $n = 5$.
Since $425{\text{Hz}} = nf$ , the fundamental frequency of the given pipe will be given by, $f = \dfrac{{425}}{n} = \dfrac{{425}}{5} = 85{\text{Hz}}$.
So the pipe will be closed at one end.
Step 2: Express the relation for the length of the closed pipe.
The fundamental frequency of a closed pipe is expressed as $f = \dfrac{v}{{4l}}$ where $v$ is the speed of sound in air and $l$ is the length of the pipe.
Then the length of a closed pipe can be expressed as $l = \dfrac{v}{{4f}}$ -------- (1)
Substituting for $v = 340{\text{m}}{{\text{s}}^{ - 1}}$ and $f = 85{\text{Hz}}$ in equation (1) we get, $l = \dfrac{{340}}{{4 \times 85}} = 1{\text{m}}$ .
$\therefore$ the length of the pipe is obtained to be 1 m.

So the correct option is A.

Note: For a closed pipe the three successive harmonics are given by, $n$ , $\left( {n + 2} \right)$ and $\left( {n + 4} \right)$ . Here $425 = nf$ , $595 = \left( {n + 2} \right)f$ and $765 = \left( {n + 4} \right)f$ . The simplified ratio of the three harmonics gives $n = 5$ as an integer. So we can confirm that the pipe is a closed one. The given harmonics are the 5th, 7th and 9th harmonics. The fundamental frequency is the lowest possible frequency and the harmonics will be multiples of this fundamental frequency.