Question 136: Waves

Question

A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities $ho_1$ and $ho_2$, respectively. The frequency of $9^{	ext {th }}$ harmonic of closed tube is identical with $4^{	ext {th }}$ harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is $ho_1: ho_2=1: 16$, then the length of the open tube is :

Accepted Answer

We know, for closed pipe,
$${f_n} = {{nv} \over {4l}},\,n = 1,3,5,7$$
for open pipe, $${f_n} = {{nv} \over {2L}},\,n = 1,2,3,4$$
So, 9$^{th}$ harmonic of closed pipe $$ = {{9{v_1}} \over {4{l_1}}}$$
4$^{th}$ harmonic of open pipe $$ = {{4{v_2}} \over {2{l_2}}} = {{2{v_2}} \over {{l_2}}}$$
$	herefore$ $${{9{v_1}} \over {4{l_1}}} = {{2{v_2}} \over {{l_2}}} \Rightarrow {{{l_2}} \over {{l_1}}} = {8 \over 9}{{{v_2}} \over {{v_1}}}$$
We know, $$v = \sqrt {{\beta  \over ho }} $$
So, $${{{v_2}} \over {{v_1}}} = \sqrt {{{{ho _1}} \over {{ho _2}}}} $$ (for same $\beta$)
hence, $${{{l_2}} \over {{l_1}}} = {8 \over 9}\sqrt {{{{ho _1}} \over {{ho _2}}}}  = {8 \over 9}\sqrt {{1 \over {16}}}  = {8 \over 9} 	imes {1 \over 4}$$
$$ \Rightarrow {l_2} = {2 \over 9} 	imes {l_1} \Rightarrow {l_2} = {2 \over 9} 	imes 10 = {{20} \over 9}cm$$

Answer

$\frac{15}{7} \mathrm{~cm}$

Answer

$\frac{20}{9} \mathrm{~cm}$

Answer

$\frac{20}{7} \mathrm{~cm}$

Answer

$\frac{15}{9} \mathrm{~cm}$

Waves

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