The closed cylinder of FIGURE P16.74 has a tight-fitting but...

The closed cylinder of FIGURE P16.74 has a tight-fitting but friction less piston of mass $M .$ The piston is in equilibrium when the left chamber has pressure $p_{0}$ and length $L_{0}$ while the spring on the right is compressed by $\Delta L.$ a. What is $\Delta L$ in terms of $p_{0}, L_{0}, A, M,$ and $k ?$ b. Suppose the piston is moved a small distance $x$ to the right. Find an expression for the net force $\left(F_{x}\right)_{\infty \text { et }}$ on the piston. Assume all motions are slow enough for the gas to remain at the same temperature as its surroundings. c. If released, the piston will oscillate around the equilibrium position. Assuming $x \ll L_{0}$ find an expression for the oscillation period $T$ Hint: Use the binomial approximation.

The closed cylinder of FIGURE P16.74 has a tight-fitting but friction less piston of mass $M .$ The piston is in equilibrium when the left chamber has pressure $p_{0}$ and length $L_{0}$ while the spring on the right is compressed by $\Delta L.$
a. What is $\Delta L$ in terms of $p_{0}, L_{0}, A, M,$ and $k ?$
b. Suppose the piston is moved a small distance $x$ to the right. Find an expression for the net force $\left(F_{x}\right)_{\infty \text { et }}$ on the piston. Assume all motions are slow enough for the gas to remain at the same temperature as its surroundings.
c. If released, the piston will oscillate around the equilibrium position. Assuming $x \ll L_{0}$ find an expression for the oscillation period $T$ Hint: Use the binomial approximation.

Key Concepts

Hooke’s Law and Spring Compression

Hooke’s Law states that the force exerted by a spring is directly proportional to its displacement from equilibrium (F = k?L). This principle is used to determine the spring force that acts against the gas pressure and piston’s weight. Establishing this relationship is important for linking the mechanical displacement of the spring with the pressure conditions in the cylinder.

Small Oscillations and Linearization via the Binomial Approximation

When the piston is displaced by a small distance, the resulting change in pressure in the gas can be approximated using a binomial expansion. This linearization simplifies the non-linear equations, allowing the system to be modeled as a simple harmonic oscillator. This concept is fundamental to finding the oscillation period by analyzing how the net restoring force varies linearly with displacement.

Equilibrium Analysis

This concept involves setting the net forces acting on the piston equal to zero. The piston, being in equilibrium, is subject to a balance between the force due to gas pressure and the mechanical forces (weight and spring force). Analyzing equilibrium is essential to relate physical quantities such as pressure, volume, and spring displacement, which sets the foundation for further dynamic analysis when the piston is displaced.

Isothermal Process and the Ideal Gas Law

In the problem’s context, the gas maintains a constant temperature, meaning that its behavior follows the isothermal process. Under this constraint, the ideal gas law (pV = constant) is used to relate changes in pressure with changes in volume when the piston moves. This concept is crucial for deriving the gas pressure variations as the chamber’s length changes, contributing to the net force on the piston.

Transcript

00:01 In this problem, we have, let's see here, we have a string.

00:08 We have a closed container here.

00:12 One side of it, there's a spring on one side of it, and then the pressure over here is zero.

00:18 So we have a vacuum over here where the spring is.

00:22 And so the pressure over here is some nominal pressure, p .o.

00:27 Now, we are asked first to what is the delta l in terms of, so here we have this initial length, and then we want to figure out what delta l is in terms of the other parameters.

00:50 So in this problem, we don't have any value, so everything is going to be parametric.

00:55 So we do a little free by diagram of this.

00:57 And on this side, we do a little free by diagram of this.

00:58 And on this side, we just have the spring force because there's no pressure.

01:03 And on this side we have the nominal for nominal the force from the pressure in here.

01:10 So that's the pressure times the cross -sectional area of this whatever this is, a rectangle or a tube or something.

01:19 And so we know the spring force is k times the change in length over here, so the displacement here.

01:30 At where at l -n, not, we're assuming that the spring is uncompressed.

01:34 Did they actually tell us that or not? when, well, the spring on the right is compressed by l.

01:48 What is it say? the piston is an equilibrium when pressure, okay, yeah, so this is an equilibrium.

01:57 So i guess i'm not, i'm not sure if delta l, if delta l is included in this.

02:06 I think it is actually.

02:09 This whole length would be, they're saying that it's in equilibrium when this length is l -0.

02:18 So delta l would be kind of included in that value.

02:25 So anyway, we can figure out just, you know, before spallance here, that delta l is p -0a over k.

02:36 Now they asked if i, suppose the piston has moved a slight distance to the right, x, by amount x, find an expression for the net force on the piston.

02:48 So assuming everything is slow so that the temperatures don't change, the temperature remains whatever it is at the surrounding temperature.

02:58 So if we perturb this over a little bit, then we get some new pressure over here.

03:09 And so the net force is going to be this new pressure, minus the spring force.

03:14 But we now, we know it's delta l the spring is now compressed delta l plus x now we can use ideal gas law so the new pressure over here times the new volume over here divided by the new temperature over here is the old pressure times the old volume times the old temperature where the temperatures are the same so we that cancels out now the new pressure we don't know the new volume we know is know is is l0 plus x.

03:52 So we had l0 here.

03:55 And so delta l was included in this.

03:59 So we now perturbited by an amount x.

04:03 So that is now, it's now l0 plus x times a cross -sectional area.

04:11 And again before the volume was l0 times a.

04:16 So what we find is that the pressure now in here is l0 divided by l0.

04:21 L0 plus x times the old pressure.

04:26 So if x is positive, then we see that this pressure has dropped, which makes sense because the volume increased.

04:37 If x is negative, then this pressure increases.

04:44 So we can then look at our net force here.

04:49 So we have p prime times a, so that finds up being l0a times divided by l -not plus x times p -not equals minus k delta l plus x.

05:11 So putting in delta l, which we know from before, and then rearranging some things, we get that the force is, why is it being x times p -not, x, p -not, a all over l0 plus x minus kx.

05:39 Okay.

05:40 Now, i believe the solutions that they have given us are, again, is kind of weird because i don't think that it could possibly be right.

05:57 They have, they have this force here does not depend on x or does not depend on i can't imagine how the k could drop out of this.

06:12 It just doesn't seem like it would be possible that the k would drop out of this.

06:17 Because, you know, if k were infinite, then this is rigid, then, i mean, you would basically just be saying, you know, you would just have a closed container here.

06:31 And so you'd have something that was, you know, was not, was not, did not compress.

06:41 At all.

06:43 So if k is infinite, then this thing isn't moving.

06:50 And so i don't know how, you know, if k was zero, then we could see, you know, that would make sense that, okay, you expand this...

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