Thermodynamics problem -- Pressure oscillations in a jar

[LagrangeEuler]

I add a Figure with the problem and solution. I have difficulty with a solution to the given problem. Why $F=-kx=Adp$, I do not understand minus sign because we are working with scalars not vectors. It is correct to say that
$\vec{F}=-kx\vec{i}$, but is not correct to say that $F=-kx$. Can you explain to me this part?

 

[BvU]

It is correct to say that $\vec{F}=-kx\vec{i}$, but is not correct to say that $F=-kx$. Can you explain to me this part?

I would prefer $\ \vec F = - k\vec x\$. The minus sign indicates that the directions are opposite. For motion along a straight line one can dispose of the vector character, but the minus sign stays:$$
\vec F = - k\vec x \Leftrightarrow F \hat \imath = - k x\hat \imath\quad \Rightarrow \quad F = - k x $$

 

[LagrangeEuler]

Is it correct to write
(p_0+\frac{mg}{A})V_0^{\gamma}=p_2(V_0-Ax)^{\gamma}?
From that
p_2=\frac{(p_0+\frac{mg}{A})V_0^{\gamma}}{(V_0-Ax)^{\gamma}}
and
p_2=\frac{p_0+\frac{mg}{A}}{(1-\frac{Ax}{V_0})^{\gamma}}
Can we assume here that $x$ is very small so that
(1+x)^n \approx 1+nx?
From that
p_2=\frac{p_0+\frac{mg}{A}}{1-\frac{\gamma A x}{V_0}}

 

[BvU]

Is it correct to write
(p_0+\frac{mg}{A})V_0^{\gamma}=p_2(V_0-Ax)^{\gamma}\ \ ?

Your book says $\ x = dV/A\$ so it chooses a coordinate system with upwards as positive x-direction.
So I would write $V_2 = V_0 + Ax$. Other than that: yes.

One small thing: x can be many $\mu$m, however. What you mean is $\displaystyle {{Ax\over V}<< 1}\$.

From your last expression (sign corrected) : $$p_2 = \left (p_0+\frac{mg}{A}\right )\ \left (1 - \frac{\gamma A^2 x}{V_0}\right )\quad \Rightarrow \quad \Delta F = p_2A-p_1A=-\frac{\gamma A^2 }{V_0}\left (p_0+\frac{mg}{A}\right )\,x$$

$\$

 

[Chestermiller]

Is it correct to write
(p_0+\frac{mg}{A})V_0^{\gamma}=p_2(V_0-Ax)^{\gamma}?
From that
p_2=\frac{(p_0+\frac{mg}{A})V_0^{\gamma}}{(V_0-Ax)^{\gamma}}
and
p_2=\frac{p_0+\frac{mg}{A}}{(1-\frac{Ax}{V_0})^{\gamma}}
Can we assume here that $x$ is very small so that
(1+x)^n \approx 1+nx?
From that
p_2=\frac{p_0+\frac{mg}{A}}{1-\frac{\gamma A x}{V_0}}

I like what you did here. But, now you can take it one step further by continuing the expansion for small x: $$p_2=(p_0+\frac{mg}{A})+(p_0+\frac{mg}{A})\frac{\gamma A x}{V_0}$$So, $$\Delta p=(p_0+\frac{mg}{A})\frac{\gamma A x}{V_0}$$So the net upward force on the ball is $$F=A\Delta p=(p_0+\frac{mg}{A})\frac{\gamma A^2 }{V_0}x$$where x is the downward displacement.

 

[LagrangeEuler]

Your book says $\ x = dV/A\$ so it chooses a coordinate system with upwards as positive x-direction.
So I would write $V_2 = V_0 + Ax$. Other than that: yes.

One small thing: x can be many $\mu$m, however. What you mean is $\displaystyle {{Ax\over V}<< 1}\$.

From your last expression (sign corrected) : $$p_2 = \left (p_0+\frac{mg}{A}\right )\ \left (1 - \frac{\gamma A^2 x}{V_0}\right )\quad \Rightarrow \quad \Delta F = p_2A-p_1A=-\frac{\gamma A^2 }{V_0}\left (p_0+\frac{mg}{A}\right )\,x$$

$\$

Thanks a lot. Could you just explain me better part $V_0-Ax$ $\rightarrow$ $V_0+Ax$? I have problem to understand why $V_2=V_0+Ax$ and relation of this and $x=dV/A$.

 

[Chestermiller]

The Taylor series expansion for $\frac{1}{1-\frac{\gamma Ax}{V_0}}$ is $\left(1+\frac{\gamma Ax}{V_0}+...\right)$

To linear terms in x, this is $$\frac{1}{1-\frac{\gamma Ax}{V_0}}\approx \left(1+\frac{\gamma Ax}{V_0}\right)$$

 

[BvU]

Thanks a lot. Could you just explain me better part $V_0-Ax$ $\rightarrow$ $V_0+Ax$? I have problem to understand why $V_2=V_0+Ax$ and relation of this and $x=dV/A$.

If the ball moves up by a distance $x$, the volume increases by $Ax\$, that's all !