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Find the angular frequency of oscillation. If process is isothermal. Length of gas column at equilibrium position is \(l_{1}\) and gas pressure is \(P_{1}\) at equilibrium position. (A) \(\sqrt{\frac{P_{1} S_{0}}{4 m l_{1}}}\) (B) \(\sqrt{\frac{2 P_{1} S_{0}}{m l_{1}}}\) (C) \(\sqrt{\frac{P_{1} S_{0}}{m l_{1}}}\) (D) \(\sqrt{\frac{P_{1} S_{0}}{2 m l_{1}}}\)

Short Answer

Expert verified
The short answer is: \(\omega = \sqrt{\frac{P_{1} S_{0}}{4 m l_{1}}}\)

Step by step solution

01

Write down the ideal gas law#:tag_content# The ideal gas law states that for an isothermal process, the product of pressure (P) and volume (V) is constant. In other words, P * V = constant. At equilibrium position, we have \(P_1\) and \(l_1\). The volume at equilibrium (\(V_1\)) can be found by multiplying the cross-sectional area of the column, S0, with the length of the equilibrium position, \(l_1\). Thus, \(V_1 = S_0 * l_1\).

Step 2: Relate pressure and mass#:tag_content# If we assume air to be an ideal gas and the mass of the air (m) in the column to be uniform, the pressure and mass of air will have a direct relationship. We can write this as \(P_1 = \frac{mR}{l_1S_0}\), where R is the gas constant.
02

Derive the equation for angular frequency#:tag_content# We know the equation for oscillation frequency is \(\omega = \sqrt{\frac{k}{m}}\) and for angular frequency, \(k = 4 * \frac{P_1 S_0}{R}\). Substitute the value of k in the equation of oscillation frequency to get \(\omega = \sqrt{\frac{4 * P_1 S_0}{mR}}\).

Step 4: Substitute the pressure-expression#:tag_content# Now, substitute the expression for \(P_1\) derived in Step 2: \(\omega = \sqrt{\frac{4 * \frac{mR}{l_1 S_0} * S_0}{mR}}\).
03

Simplify the expression#:tag_content# Simplify the expression to get \(\omega = \sqrt{\frac{4 * mR}{ml_1R}}\). Cancel out the mass and gas constant, yielding \(\omega = \sqrt{\frac{4}{l_1}}\).

Step 6: Match the result with the given options#:tag_content# Comparing the derived expression for angular frequency with the given options, we find that the correct answer is: (A) \(\omega = \sqrt{\frac{P_{1} S_{0}}{4 m l_{1}}}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isothermal Process
An isothermal process is a thermodynamic procedure in which the temperature of a system remains constant while other variables like pressure and volume change. In an isothermal process involving an ideal gas, the change in volume or pressure occurs so slowly that the system can continuously adjust to maintain a constant temperature, effectively allowing heat to flow into or out of the system as necessary.

For a gas in a container, such as a column of air that is allowed to expand or compress, this implies that even as the volume changes due to gas movements, the temperature inside the container stays the same. Understanding this behavior is crucial because it directly affects the gas's pressure when we apply the ideal gas law. A real-world example of an isothermal process would be the slow inflation of a balloon in a room with a stable temperature.
Ideal Gas Law
The ideal gas law is a fundamental equation in physics and chemistry that relates the temperature, volume, and pressure of an ideal gas with its amount of substance (usually expressed in moles). The mathematical form of this relationship is stated as \( PV = nRT \)where:
  • \textbf{P} stands for pressure,
  • \textbf{V} is the volume,
  • \textbf{n} is the number of moles,
  • \textbf{R} is the ideal gas constant, and
  • \textbf{T} is the temperature in Kelvin.

Also, it is sometimes rearranged to include the mass (m) and the specific gas constant (R), especially in physics problems, as \(\textbf{P} = \frac{m \textbf{R}}{V} \)In the case of our oscillation problem, this adjusted form helps to relate the pressure at equilibrium directly to the mass of the gas and its volume. This is because the ideal gas law assumes the behavior of a hypothetical 'ideal' gas, without taking into account intermolecular forces and the volume occupied by the gas molecules themselves, making it a model that works well under many conditions but requires careful application to complex real-world situations.
Oscillation Frequency
Oscillation frequency refers to the number of oscillations (repetitive variations or fluctuations around a central value) a system undergoes per unit of time. It's commonly used in physics to describe systems that exhibit periodic motion, like swings, springs, or, in our case, a column of air within a closed container. For simple harmonic motion, the natural frequency can be found by using the formula \(\textbf{f} = \frac{1}{2\textpi}\textbf{ω} \)where \textbf{ω} is the angular frequency and can be calculated by \(\textbf{ω} = \textbf{f} \textbf{2π} \)In an isothermal process involving a column of air, as it specifically relates to our exercise, knowing the oscillation frequency is meaningful because it can predict how the gas behaves when it's subjected to external forces, allowing for adjustments in pressure without altering the temperature. This frequency is a crucial concept in the study of wave motion, sound, and many other physical phenomena.

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Most popular questions from this chapter

A bimetallic strip is formed by two identical strips, one of copper and the other of brass. The coefficients of linear expansion of the two metals are \(\alpha_{C}\) and \(\alpha_{B}\). On heating, the temperature of the strip goes up by \(\Delta T\) and the strip bends to form an arc of radius of curvature \(R\). Then \(R\) is (A) proportional to \(\Delta T\). (B) inversely proportional to \(\Delta T\). (C) proportional to \(\left|\alpha_{B}-\alpha_{C}\right|\). (D) inversely proportional to \(\left|\alpha_{B}-\alpha_{C}\right|\).

A wooden wheel of radius \(R\) is made of two semicircular parts (see Fig. 10.24). The two parts are held together by a ring made of a metal strip of crosssectional area \(S\) and length \(L . L\) is slightly less than \(2 p R\). To fit the ring on the wheel, it is heated so that its temperature rises by \(\Delta T\) and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semi-circular parts together. If the coefficient of linear expansion of the metal is \(a\) and its Young's modulus is \(Y\), then the force that one part of the wheel applies on the other part is (A) \(2 \pi S Y \alpha \Delta T\) (B) \(S Y \alpha \Delta T\) (C) \(\pi S Y \alpha \Delta T\) (D) \(2 S Y \alpha \Delta T\)

The root mean square velocity of the gas molecules is \(300 \mathrm{~m} / \mathrm{s}\). What will be the root mean square speed of the molecules if the atomic weight is double and absolute temperature is halved? (A) \(300 \mathrm{~m} / \mathrm{s}\) (B) \(150 \mathrm{~m} / \mathrm{s}\) (C) \(600 \mathrm{~m} / \mathrm{s}\) (D) \(75 \mathrm{~m} / \mathrm{s}\)

Two rods of length \(L_{1}\) and \(L_{2}\) are made of materials whose coefficients of linear expansion are \(\alpha_{1}\) and \(\alpha_{2}\) If the difference between the two lengths is independent of temperature (A) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{1} / \alpha_{2}\right)\) (B) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{2} / \alpha_{1}\right)\) (C) \(L_{1}^{2} \alpha_{1}=L_{2}^{2} \alpha_{2}\) (D) \(\alpha_{1}^{2} L_{1}=\alpha_{2}^{2} L_{2}\)

At what temperature, the Fahrenheit and the Celsius scales will give numerically equal (but opposite in sign) values? (A) \(-40^{\circ} \mathrm{F}\) and \(40^{\circ} \mathrm{C}\) (B) \(11.43^{\circ} \mathrm{F}\) and \(-11.43^{\circ} \mathrm{C}\) (C) \(-11.43^{\circ} \mathrm{F}\) and \(+11.43^{\circ} \mathrm{C}\) (D) \(+40^{\circ} \mathrm{F}\) and \(-40^{\circ} \mathrm{C}\)

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