Question

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

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

Answer 1

Answer: (B)

Given, speed of sound in air $\left(v_{0}\right)=340\, ms ^{-1}$
Also, successive frequencies $425,595 $ and $765 \,Hz$ are odd frequencies. Thus given organ pipe is "closed organ pipe".
$\therefore $ Fundamental frequency
$v_{0} =\frac{595-425}{2} $
$=\frac{765-595}{2}=85\, Hz $
We know that, for a closed organ pipe,
Fundamental frequency is given by
$ v_{0} =\frac{v_{0}}{4 l}$
$\Rightarrow 85 =\frac{340}{4 \times 1} $
$\Rightarrow l =\frac{340}{4 \times 85} $
$\Rightarrow l =1 \,m $