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Q.
The fundamental frequency of a closed organ pipe is 'n'. Its length is doubled and radius is halved. Its frequency will become nearly.
Oscillations
Solution:
Fundamental frequency of the closed end organ pipe is
$v_{0 c}=\frac{v}{4(1+0.6 r)}$
After its length is doubled and radius is halved,
the fundamental frequency becomes
$v_{0 c}'=\frac{v}{4(2 l+0.6 r / 2)}$
Thus, $\frac{v_{0 c}}{v_{0 c}'}=\frac{4(21+0.6 r / 2)}{4(1+0.3 r )}$
Since, $1 >>r$
$\frac{v_{0 c}}{v_{0 c}'}=2$
$v_{0 c}'=n / 2$