Question

A student is performing the experiment of a resonance column. The diameter of the column tube is $4cm$. The frequency of the tuning fork is $512\,Hz$. The air temperature is ${{38}^{o}}C$ in which speed of sound is $336m\,{{s}^{-1}}$. The zero of the meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is-
(A). $14cm$
(B). $15.2cm$
(C). $16.4cm$
(D). $17.6cm$

Answer 1

The resonance column apparatus consists of a jar containing water and a vertical tube which is immersed in water to increase or decrease its length that is in play. The conditions of resonance vary if the tube is open or closed at one end. Check if there is error correction or not.

Formulas Used:
$L+e=\dfrac{\lambda }{4}$

Complete step by step solution:
In a resonance column experiment, the apparatus consists of a jar containing water and a vertical tube held above the jar. When the jar is immersed in water, it is closed from one end. When a vibrating tuning fork is brought near it, the waves will get reflected on the water surface and form standing waves where the nodes are formed on the water surface and antinodes are formed at the open surface.
We can adjust the length of the tube by immersing it in water accordingly. For the first resonance, the conditions are-
$L+e=\dfrac{\lambda }{4}$ ………………...(1)
Here,$L$ of the tube in which standing waves are formed
$e$ is error correction in its length
$\lambda$ is the wavelength of vibrations
Given, speed of sound in air =$336m\,{{s}^{-1}}$, frequency, $\nu =512Hz$
We know that,
$v=\nu \lambda$
In the above equation, we substitute the given values to get,
$\begin{aligned} & \Rightarrow 336=512\times \lambda \\\ & \Rightarrow \lambda =\dfrac{512}{336} \\\ & \therefore \lambda =\dfrac{16}{21} \\\ \end{aligned}$
Substituting the value of $\lambda$ in eq (1), we get,
$\begin{aligned} & L+e=\dfrac{\lambda }{4} \\\ & \Rightarrow L+e=\dfrac{16}{4\times 21} \\\ \end{aligned}$
$\Rightarrow L=\dfrac{4}{21}-e$ …………………... (2)
$e$ is given as-
$e=0.3\times D$
Here, $D$ is the diameter of the tube
$\begin{aligned} & e=0.3\times D \\\ & \Rightarrow e=0.3\times 4 \\\ & \therefore e=1.2cm \\\ \end{aligned}$
When we substitute the value of $e$ in eq (2), we get,
$\begin{aligned} & L=\dfrac{4\times {{10}^{2}}}{21}-1.2 \\\ & \Rightarrow L=19.04-1.2 \\\ & \therefore L=17.6cm \\\ \end{aligned}$

Therefore, the length of the tube is $17.6cm$. Hence, the correct option is (D).

Note:
Resonance frequency is the frequency at which a vibrating object vibrates at higher amplitude. In the resonance column experiment, the condition for resonance is when the air particles vibrate with the same frequency as the tuning fork. The point on a standing wave where the amplitude is minimum is called a node, while the point on the standing wave where the amplitude is maximum is called an antinode.