STEM Problem Detail

Problem Statement

Prompt

Provide the step by step solution of Ex. 20.

Solution Rendering...

**Solution**

Here the motion of the fluid takes place in such a manner that each element
of the fluid moves towards the centre. Hence the free surface would be
spherical. Thus the fluid velocity $$v$$ will be radial and will be a function
of $$x$$ (the radial distance from the centre of the spherical shell) and $$t$$
only. Let $$p$$ be the pressure at a distance $$x$$.

------------------------------------------------

### Equation of Continuity

\[
x^2 v = F(t)
\]

so that
\[
\frac{\partial v}{\partial t} = \frac{F'(t)}{x^2}.
\tag{1}
\]

------------------------------------------------

### Equation of Motion

\[
\frac{\partial v}{\partial t}
+ v\frac{\partial v}{\partial x}
= -\frac{1}{\rho}\frac{\partial p}{\partial x}
\]

or, using (1),
\[
\frac{F'(t)}{x^2}
+ \frac{\partial}{\partial x}\!\left(\frac{1}{2}v^2\right)
= -\frac{1}{\rho}\frac{\partial p}{\partial x}.
\]

Integrating with respect to $$x$$,
\[
-\frac{F'(t)}{x}
+ \frac{1}{2}v^2
= -\frac{p}{\rho} + C.
\tag{2}
\]

------------------------------------------------

### Mass Conservation

Let $$r$$ and $$R$$ be the internal and external radii of the shell at any time $$t$$.
Since the shell contains gas, there is pressure on the inner surface.
Let $$p=p_1$$ when $$x=r$$.

Since the total mass of the liquid is constant,
\[
\left(\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3\right)\rho
=
\left(\frac{4}{3}\pi b^3 - \frac{4}{3}\pi a^3\right)\rho,
\]

or
\[
R^3 - r^3 = b^3 - a^3.
\tag{3}
\]

------------------------------------------------

### Boyle's Law

Since the initial pressure of the gas is equal to atmospheric pressure $$\Pi$$,
Boyle's law gives
\[
\frac{4}{3}\pi r^3 p_1 = \frac{4}{3}\pi a^3 \Pi,
\]

so that
\[
p_1 = \frac{a^3 \Pi}{r^3}.
\tag{4}
\]

------------------------------------------------

### Boundary Conditions

Since the outer surface is exposed to atmosphere, when $$x=R$$,
\[
v=\frac{dR}{dt}=U,
\qquad
p=\Pi.
\]

Substituting in (2),
\[
-\frac{F'(t)}{R}
+ \frac{1}{2}U^2
= -\frac{\Pi}{\rho} + C.
\tag{5}
\]

Again, when $$x=r$$, $$v=\frac{dr}{dt}=u$$ and $$p=p_1$$,
\[
-\frac{F'(t)}{r}
+ \frac{1}{2}u^2
= -\frac{a^3\Pi}{\rho r^3} + C.
\tag{6}
\]

------------------------------------------------

### Elimination of Constant

Subtracting (6) from (5),
\[
\left(\frac{1}{r}-\frac{1}{R}\right)F'(t)
+ \frac{1}{2}(U^2-u^2)
=
\frac{\Pi}{\rho}\left(\frac{a^3}{r^3}-1\right).
\tag{7}
\]

------------------------------------------------

### Small Oscillations

For small oscillations, $$u^2$$ and $$U^2$$ are negligible. Hence,
\[
F'(t)
=
\frac{\Pi}{\rho}
\frac{a^3-r^3}{r^2}
\frac{R}{R-r}.
\tag{8}
\]

------------------------------------------------

### Final Reduction

From continuity,
\[
F(t)=r^2u.
\]

Thus,
\[
F'(t)=2ru^2+r^2\frac{d^2r}{dt^2}.
\]

Neglecting $$u^2$$,
\[
F'(t)=r^2\frac{d^2r}{dt^2}.
\]

Substituting in (8),
\[
r^2\frac{d^2r}{dt^2}
=
\frac{\Pi}{\rho}
\frac{a^3-r^3}{r^2}
\frac{R}{R-r}.
\tag{9}
\]

Let small displacements be
\[
r=a+x,
\qquad
R=b+x'.
\tag{10}
\]

Then,
\[
\frac{d^2r}{dt^2}=\ddot{x}.
\]

Equation (9) becomes
\[
\ddot{x}
=
\frac{\Pi}{a\rho}
\frac{1-(1+x/a)^3}{(1+x/a)^4}
\frac{b+x'}{x'-x+b-a}.
\tag{11}
\]

Using (10) in (3),
\[
(b+x')^3-(a+x)^3=b^3-a^3.
\]

To first order,
\[
3b^2x'-3a^2x=0,
\qquad
x'=\frac{a^2}{b^2}x.
\tag{12}
\]

Substituting in (11) and neglecting higher powers,
\[
\ddot{x}
=
-\frac{3b\Pi}{a^2\rho(b-a)}x.
\]

Hence the motion is simple harmonic.

------------------------------------------------

### Time Period

\[
T
=
\frac{2\pi}{\sqrt{\dfrac{3b\Pi}{a^2\rho(b-a)}}}
=
2\pi a
\left(\frac{\rho(b-a)}{3b\Pi}\right)^{1/2}.
\]

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Problem #31