Resonance Column Method - Practice Questions & MCQ

Resonance column method

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

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

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

Let at this position the length of 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,

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

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