Question

Four students perform a resonance tube experiment. The first, second and the third resonant length observed are respectively ${{{l}}_{{1}}}{{,}}{{{l}}_{{2}}}$ and ${{{l}}_{{3}}}$. Whose readings are most appropriate?
A) ${{{l}}_{{1}}}{{ = 20 cm, }}{{{l}}_{{2}}}{{ = 58 cm, }}{{{l}}_{{3}}}{{ = 98 cm}}$
B) ${{{l}}_{{1}}}{{ = 20 cm, }}{{{l}}_{{2}}}{{ = 58 cm, }}{{{l}}_{{3}}}{{ = 96 cm}}$
C) ${{{l}}_{{1}}}{{ = 20 cm, }}{{{l}}_{{2}}}{{ = 62 cm, }}{{{l}}_{{3}}}{{ = 104 cm}}$
D) ${{{l}}_{{1}}}{{ = 20 cm, }}{{{l}}_{{2}}}{{ = 60 cm, }}{{{l}}_{{3}}}{{ = 100 cm}}$

Answer 1

Hint: The principle of resonance tube gives us ways to vary the length of the closed tube. Here, the water column in a glass tube acts as a piston to maintain a column of air of the desired length. However, the length can be varied simply by moving the reservoir up and down to the desired height. The most appropriate readings can be calculated by forming an equation for different lengths of resonance.

Complete step by step solution:
The resonance tube is open at one end and the fundamental frequency is given by
${{f = }}\dfrac{{{v}}}{{{{4 }}{{{l}}_1}}}$
On rearranging the terms
\[{{{l}}_{{1}}}{{ = }}\dfrac{{{{{v}}_{{0}}}}}{{{{4 f}}}}\]
For first resonance, length is just half loop
Let ${{e }}$ be end correction
For first resonance length
$\Rightarrow {{{l}}_{{1}}}{{ + e = }}\dfrac{{{{{v}}_{{0}}}}}{{{{4f}}}}...{{(i)}}$
For second length
$\Rightarrow {{{l}}_2}{{ + e = }}\dfrac{{{{3}}{{{v}}_{{0}}}}}{{{{4f}}}}...{{(ii)}}$
For third length
$\Rightarrow {{{l}}_3}{{ + e = }}\dfrac{{{{5}}{{{v}}_{{0}}}}}{{{{4f}}}}...{{(iii)}}$
Solving these three equations, in order to get the exact length of the resonance.
Multiplying equation (ii) by
$\Rightarrow {{2}}{{{l}}_2}{{ + 2e = }}\dfrac{{{{6}}{{{v}}_{{0}}}}}{{{{4f}}}}$
Subtracting the above equation by (i), we get
$\Rightarrow {{2}}{{{l}}_2}{{ + 2e - }}{{{l}}_1}{{ - e = }}\dfrac{{{{6}}{{{v}}_{{0}}}}}{{{{4f}}}}{{ }} - {{ }}\dfrac{{{{{v}}_{{0}}}}}{{{{4f}}}}$
On further solving, we get
$\Rightarrow {{2}}{{{l}}_2}{{ - }}{{{l}}_1}{{ + e = }}\dfrac{{{{5}}{{{v}}_{{0}}}}}{{{{4f}}}}$
As you can see that RHS of above equation and equation (iii) are equal, so their LHS must be equal
$\Rightarrow {{2}}{{{l}}_{{2}}}{{ - }}{{{l}}_{{1}}}{{ + e = }}{{{l}}_{{3}}}{{ + e }}$
$ \therefore {{2}}{{{l}}_{{2}}}{{ - }}{{{l}}_{{1}}} = {{{l}}_{{3}}}...{{(iv)}}$
Now from equation (iv), it is very clear that ${{{l}}_{{3}}}{{ > }}{{{l}}_{{2}}}$ and ${{{l}}_{{1}}}$
As twice of ${\operatorname{l} _2}$ when subtracting from ${{{l}}_1}$ gives ${{{l}}_{{3}}}$
So ${{{l}}_{{1}}}{{ = 20 cm, }}{{{l}}_{{2}}}{{ = 58 cm, }}{{{l}}_{{3}}}{{ = 96 cm}}$

Thus, option (B) is the correct choice.

Note: As there are three equations, in which 4 variables are there namely ${{{l}}_{{1}}}{{, }}{{{l}}_{{2}}}{{, }}{{{l}}_{{3}}}$ and $e$. Try to solve the equations in such a way that the minimum number of variables are left and there is a meaningful relationship between the left variables. Then by hit and trial method from the given equations, we can find out the correct option as there are many variables. Conclusion: twice of ${\operatorname{l} _2}$ when subtracting from ${{{l}}_1}$ gives ${{{l}}_{{3}}}$ for this only option (B) holds good.