Question

Two closed organ pipe $\mathrm{A}$ and $\mathrm{B}$ have the same length. $\mathrm{A}$ is wider than $\mathrm{B}$. They resonate in the fundamental mode at frequencies ${\mathrm{v}}_{\mathrm{A}}$ and ${\mathrm{v}}_{\mathrm{B}}$ respectively, then

• A. ${\mathrm{v}}_{\mathrm{A}}={\mathrm{v}}_{\mathrm{B}}$
• B. ${\mathrm{v}}_{\mathrm{A}}>{\mathrm{v}}_{\mathrm{B}}$
• C. ${\mathrm{v}}_{\mathrm{A}}<{\mathrm{v}}_{\mathrm{B}}$
• D. either (2) or (3) depending on the ratio of their diameters

Answer 1

The correct option is C

${\mathrm{v}}_{\mathrm{A}}<{\mathrm{v}}_{\mathrm{B}}$

Step 1: Given data

Two closed organ pipe $\mathrm{A}$ and $\mathrm{B}$ have the same length. $\mathrm{A}$ is wider than $\mathrm{B}$.

Step 2: Concept to be used

The lowest part of a harmonic vibration or the lowest frequency at which an oscillation occurs is called the fundamental mode of vibration.

Step 3: Find which pipe is the greater one

Let

$\mathrm{l}=$ length of the pipe

$\mathrm{d}=$ diameter of the pipe

The value of end correction $\mathrm{e}$ is $0.6\mathrm{r}$ for a closed organ pipe and $1.2\mathrm{r}$ for an open organ pipe, where $\mathrm{r}$ is the radius of the pipe.

In closed organ pipe.

First resonance occurs at $\frac{\mathrm{\lambda }}{4}$ .
So, in the fundamental mode of vibration of an organ pipe,
$\frac{\mathrm{\lambda }}{4}=(\mathrm{l}+0.3\mathrm{d})$
where $0.3\mathrm{d}$ is necessary end correction.
Frequency of vibration,
$\frac{\mathrm{v}}{\mathrm{\lambda }}=\frac{\mathrm{v}}{4+\left(\mathrm{l}+0.3\mathrm{d}\right)}$​
As $\mathrm{l}$ is same, wider pipe A will resonate at a lower frequency, i.e.,${\mathrm{v}}_{\mathrm{A}}<{\mathrm{v}}_{\mathrm{B}}$

Hence, option $\mathrm{C}$ is the correct answer.