Resonance Column Method MCQ - Practice Questions with Answers

Resonance column method

In this, the equipment used is a resonance tube. This apparatus is used to determine the velocity of sound in air and to compare the frequency of two turning forks.

It is a closed organ pipe with a variable length of air column. When we brought a turning fork near it, its air column vibrated with the frequency of the fork. The length of the air column varied until the frequency of the fork and the air column became equal. When frequency becomes equal, the column resonates and the note becomes loud. 

It is the full setup of the resonance tube. If a tuning fork of known frequency  is struck on a rubber pad and brought near the open end. Because of this, the air column starts oscillating. This air column behaves as a closed organ pipe and the water level is a closed end. We decrease the water level gradually and as the water level reaches a position where there is a node of the corresponding stationary wave, in the air column, resonance takes place. At this place intensity of sound will be maximum. 

Let at this position the length of the air column is $l_1$. By further decreasing water level again after some distance maximum intensity of sound is obtained where the node is obtained. Let this level is $l_2$.

If $l_1$ and $l_2$ are the length of first and second resonance, then -

$$
\begin{aligned}
& l_1+e=\frac{\lambda}{4} \text { and } l_2+e=\frac{3 \lambda}{4} \\
& \text { so, } \lambda=2\left(l_2-l_1\right)
\end{aligned}
$$
Speed of sound in air at room temperature $v=n \lambda=2 n\left(l_2-l_1\right)$

Also,

$$
\begin{gathered}
\\
\\
\Rightarrow l_2+e \\
l_1+e
\end{gathered}=3=3 l_1+2 e
$$

So, the second resonance is obtained at a length more than thrice the length of first resonance.