Question

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

• A. n/2
• B. n/3
• C. n
• D. 2n

Answer 1

Answer: (A)

n/2

Fundamental frequency of the closed end organ pipe is
${\nu }_{0c}=\frac{v}{4(l+0.6r)}$

After its length is doubled and radius is halved, the fundamental frequency becomes
${\nu }_{0c}^{\prime }=\frac{v}{4(2l+0.6r/2)}$

Thus, $\frac{{\nu }_{0c}}{{\nu }_{0c}^{\prime }}=\frac{4(2l+0.6r/2)}{4(l+0.3r)}$

Since, $l>>r$
$\frac{{\nu }_{0c}}{{\nu }_{0c}^{\prime }}=2$

${\nu }_{0c}^{\prime }=n/2$