Question

The fundamental frequency of a closed organ pipe is 'n'. Its length is doubled and radius is halved. Its frequency will become nearly.

• A. $\frac{n}{2}$
• B. $\frac{n}{3}$
• C. n
• D. 2n

Answer 1

Answer: (A)

Fundamental frequency of the closed end organ pipe is
$v_{0c}=\frac{v}{4(1+0.6r)}$
After its length is doubled and radius is halved,
the fundamental frequency becomes
$v_{0c}^{\prime }=\frac{v}{4(2l+0.6r/2)}$
Thus, $\frac{v_{0c}}{v_{0c}^{\prime }}=\frac{4(21+0.6r/2)}{4(1+0.3r)}$
Since, $1>>r$
$\frac{v_{0c}}{v_{0c}^{\prime }}=2$
$v_{0c}^{\prime }=n/2$