Question

For a certain organ pipe, three successive resonance frequencies are observed at $425,595$ and $765Hz$, respectively. The length of the pipe is (speed of sound in air $=340m{s}^{-1}$ )

• A. 0.5 m
• B. 1 m
• C. 1.5 m
• D. 2 m

Answer 1

Answer: (B)

Given, speed of sound in air $\left(v_0\right)=340m{s}^{-1}$
Also, successive frequencies $425,595$ and $765Hz$ are odd frequencies. Thus given organ pipe is "closed organ pipe".
$\therefore$ Fundamental frequency
$v_0=\frac{595-425}{2}$
$=\frac{765-595}{2}=85Hz$
We know that, for a closed organ pipe,
Fundamental frequency is given by
$v_0=\frac{v_0}{4l}$
$\Rightarrow 85=\frac{340}{4\times 1}$
$\Rightarrow l=\frac{340}{4\times 85}$
$\Rightarrow l=1m$