Question

In a resonance tube, closed at one end by a smooth moving piston and the other end open, exhibits the first three resonance lengths of air column $L_1,L_2$ and $L_3$ for the same tuning fork. Then they are related by:

• A. $L_3-L_2=L_2-L_1$
• B. $L_3=\frac{5}{3}{L}_2=5L_1$
• C. $L_3-L_2=\frac{L_2-L_1}{2}$
• D. $L_3-L_2=2(L_2-L_1)$

Answer 1

The correct option is A $L_3-L_2=L_2-L_1$

The first three resonance length of air column is $L_1,L_2$ and $L_3$

In case of resonance condition, the length of tube $L$ is satisfied

$L_n=\frac{1}{4}(2n+1)\lambda$

So,

$\begin{array}{c}{L}_1=\frac{3}{4}\lambda \\ L_2=\frac{5}{4}\lambda \\ L_3=\frac{7}{4}\lambda \\ L_3-L_2=\frac{7}{4}\lambda -\frac{5}{4}\lambda =\frac{1}{4}\left(7-5\right)\lambda =\frac{1}{2}\lambda ....\left(1\right)\\ L_2-L_1=\frac{5}{4}\lambda -\frac{3}{4}\lambda =\frac{1}{4}\left(5-3\right)\lambda =\frac{1}{2}\lambda ....\left(2\right)\end{array}$

So, by equation $\left(1\right)$ and $\left(2\right)$

$L_3-L_2=L_2-L_1$

Hence, Option $A$ is the correct answer.