Animated Solution for Physics - Waves: A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air column is the second resonance. Then,

Have you ever blown across the top of an empty glass bottle and heard a beautiful, deep hum? What you are experiencing is the magic of acoustic resonance.

In the physics laboratory, we study this exact phenomenon using a resonance column apparatus. This setup consists of a vertical tube filled with water, where the water level can be adjusted to change the length of the air column inside. By vibrating a tuning fork of a known frequency above the open end of the tube, we can establish standing waves in the air column.

This classic JEE Advanced problem tests our fundamental understanding of the boundary conditions, physical dimensions, and energy dissipation mechanisms in this experiment.

Let's break down the physics of each option to uncover the truth.

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Let's first address the physical properties and orientation of our source of sound: the tuning fork.

Sound waves propagating inside the vertical resonance tube are longitudinal waves. In a longitudinal wave, the air molecules oscillate back and forth parallel to the direction of wave propagation. Since the tube is vertical, the air molecules must vibrate vertically to set up a standing wave.

To excite these vertical oscillations efficiently, the prongs of the tuning fork must vibrate up and down. This requires the prongs to be held in a vertical plane directly above the open mouth of the tube.

If the prongs were kept in a horizontal plane, they would vibrate horizontally, creating transverse disturbances that do not couple well with the vertical air column. Therefore, Option B is incorrect.

Now, let's think about the physical scale of a tuning fork's vibration. When you strike a tuning fork, you can hear it clearly, but can you see the prongs moving by a whole centimeter? Absolutely not!

A displacement of $1\text{ cm}$ is massive for a stiff metal prong. If a tuning fork were to vibrate with such a large amplitude, the mechanical stress would exceed the elastic limit of the steel or aluminum, causing the fork to permanently deform or snap.

In reality, the amplitude of vibration of a standard tuning fork is extremely small, typically in the range of $0.1\text{ mm}$ to $1\text{ mm}$. Thus, Option C is incorrect.

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Now, let's dive into the core mathematical beauty of standing waves in pipes: end correction.

We often teach that the open end of a pipe is a displacement antinode because the air molecules there are completely free to move. While this is a good approximation, it is not perfectly true.

The air molecules just outside the open mouth of the tube also get pushed and pulled by the vibrating air column inside. Because this external air also has mass and inertia, the pressure does not drop to atmospheric pressure exactly at the physical boundary of the tube. Instead, the pressure boundary (and thus the displacement antinode) lies slightly outside the tube at a distance $e$, known as the end correction.

For a cylindrical tube of radius $r$, the end correction is given by:

For the first resonance, the air column contains a single node at the water surface and a single antinode slightly above the open end. The distance between a node and its consecutive antinode is exactly one-quarter of the wavelength ($\frac{\lambda }{4}$).

Therefore, the effective length of the resonating air column is:

Solving for the physical length of the air column $l_1$, we get:

Since the radius $r>0$, the end correction $e$ is a positive quantity. This means:

Thus, the physical length of the air column at the first resonance is indeed somewhat shorter than $\frac{\lambda }{4}$. This makes Option D absolutely correct.

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Finally, let's compare the loudness (intensity) of the sound heard at the first and second resonances.

When the air column resonates, the air molecules oscillate vigorously. As they slide against the inner walls of the glass tube, they experience viscous drag. Additionally, the rapid compressions and rarefactions of the air lead to local temperature changes, causing thermal energy to conduct into the tube walls.

Both of these mechanisms—viscous friction and thermal conduction—dissipate acoustic energy into heat.

At the second resonance, the length of the air column is approximately three times longer than at the first resonance ($l_2\approx 3l_1$). A longer air column means: 1. A much larger surface area of the tube wall is in contact with the vibrating air, leading to significantly higher viscous losses. 2. A larger volume of air is oscillating, leading to greater bulk thermal dissipation.

Because of this increased damping, the acoustic energy radiated out of the tube is lower at the second resonance. Consequently, the sound heard at the first resonance is noticeably louder and more intense than at the second resonance. Therefore, Option A is correct.

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By analyzing the physical and mathematical principles of the resonance column experiment, we have established that: - Option A is correct because viscous and thermal damping are lower in the shorter air column of the first resonance. - Option D is correct because the end correction shifts the antinode slightly outside the tube, making the physical column length shorter than $\frac{\lambda }{4}$.