Problem 43 - Physics

Question

A vertical tube is closed at one end and open to air at the other end. The air pressure is \(1.01 \mathrm{X} 10^{5} \mathrm{~Pa}\). The tube has a length of \(0.75 \mathrm{~m}\). Mercury (mass density \(=13\) \(\left.600 \mathrm{~kg} / \mathrm{m}^{3}\right)\) is poured into it to shorten the effective length for standing waves. What is the absolute pressure at the bottom of the mercury column, when the fundamental frequency of the shortened, air-filled tube is equal to the third harmonic of the original tube?

Step-by-Step Solution

Answer

The absolute pressure at the bottom of the mercury column is 103,334 Pa.

Step 1: Understand the Problem

The problem involves a tube where initially the air column can produce standing waves. It becomes shortened by adding mercury. We need to find the absolute pressure at the bottom of the mercury column when the fundamental frequency equals the tube's third harmonic frequency before mercury was added.

Step 2: Determine the Fundamental and Harmonic Conditions

Initially, consider the open tube's harmonic conditions. For an open-closed tube, the fundamental frequency has a quarter wavelength equal to the tube length. When the third harmonic (3/4 wavelength) frequency condition matches, the length of the air column must adapt for the shortened frequency.

Step 3: Calculate the Full Tube's Fundamental and 3rd Harmonic

Using the formula for the speed of sound, \(v = f \lambda\), solve \(f_3 = 3(f_1)\), where \(f_1\) is the fundamental frequency of 0.75m. The third harmonic \(\lambda_3 = \frac{4}{3}L\). Equate \(\frac{v}{\lambda_3} = 3\times \frac{v}{4L}\); hence, \(f_3 = \frac{3v}{4L}\).

Step 4: Calculate the Frequency of the Shortened Tube

For the shortened tube, the length is no longer \(L\), instead a new length \(L'\) must create a fundamental condition matching the third harmonic of the original length. Find \(L'\) using \(f' = \frac{v}{4L'} = f_3\).

Step 5: Determine Shortened Length Due to Mercury

The shortened air column length is determined by subtracting the height of the mercury column from the tube: \(L' = L - h_M\). From previous steps, you can find \(h_M = \frac{1}{3}L\).

Step 6: Calculate Mercury Height and Pressure

With \(h_M = \frac{0.75}{3} = 0.25m\), the pressure due to mercury is \(P_M = \rho gh_M\) where \(\rho = 13,600 kg/m^3\) and \(g = 9.81 m/s^2\). The mercury pressure is 3,334 Pa.

Step 7: Calculate Absolute Pressure at Bottom

Add the atmospheric pressure to the mercury pressure: \(P_{absolute} = P_{atm} + P_M = 1.01 \times 10^5 + 3,334\). Resulting in \(103,334\) Pa.

Key Concepts

Standing Waves

Standing waves are a fascinating phenomenon, particularly common in musical instruments and fluid dynamics problems like this one. These occur when two waves of the same frequency and amplitude travel in opposite directions and interfere with each other. In a tube closed at one end and open at the other, standing waves develop under specific conditions.
For such tubes, the simplest standing wave (the fundamental frequency) occurs when the air column's length is one-quarter of the wavelength of the sound wave. Various harmonics follow, with the third harmonic having three-quarters of the wavelength fitting into the tube's length.
Understanding standing waves helps us determine how sound behaves in a tube, significantly influencing the way we calculate sound frequencies, as in this exercise.

Harmonics

Harmonics play a key role in the analysis of sound within a tube system. They are multiples of the fundamental frequency, with each harmonic representing a mode in which the air column can naturally vibrate.
In our exercise, we deal with the third harmonic. For a tube closed at one end (like our original condition), the third harmonic requires that three-quarters of the sound wave fits into the tube. This correlation between harmonics and the tube’s dimensions allows us to set boundary conditions for calculating exact sound frequencies.
This means once we know the fundamental frequency, calculating any harmonic frequency becomes straightforward by multiplying the harmonic number by the fundamental frequency.

Pressure Calculation

Pressure calculation in a fluid-filled tube involves considering both atmospheric pressure and the pressure exerted by the fluid column. In this example, mercury is the fluid that affects how pressure is distributed within the tube.
The mercury's column height ( h_M) shortens the air column, influencing the standing wave conditions. We calculated it using the differential pressure formula, \( P_M = \rho gh_M \), where \( \rho \) is the density of mercury and \( g \) is the gravitational acceleration.
Adding atmospheric pressure to this pressure gives the absolute pressure at the bottom of the tube, ensuring that we get a complete measure that incorporates both environmental and mechanical contributions.

Fluid Dynamics

Fluid dynamics is essential in explaining how mercury's introduction alters wave behavior in the tube. It studies how forces interact with fluid substances like mercury, affecting static and dynamic pressure conditions.
By pouring mercury into the tube, we effectively change the tube's length that resonates. The fundamental frequency adjusts due to this new configuration, demonstrating how fluid presence can change acoustic properties.
This principle of fluid dynamics helps us understand not only this experiment’s setup but also broader applications in engineering and natural phenomena, where liquid environments alter sound wave travel.