Chapter 19.3, Problem 19.98P

To determine

(a)

The period of vibration of the shell when it is displaced vertically and released.

Answer to Problem 19.98P

The period of vibration: ${\tau }_n=0.352s$

Explanation of Solution

Given information:

Mass of the hollow-spherical shell is 500-g.

Radius of the spherical shell is 80 mm.

Constant of the spring is 500 N/m.

Calculations:

For the two positions as shown in the figure:

$\begin{array}{l}\text{At Position 2:}\\ \\ \text{Kinetic energy: }{T}_2=0\\ \text{Potential energy: }{V}_2=\frac{1}{2}k{x}_m^2\end{array}$

$\begin{array}{l}\text{At Position 1 :}\\ \text{Kinetic energy: }{T}_1=T_{spere}+T_{fluid}=\frac{1}{2}{m}_s{v}_m^2+\frac{1}{4}\rho V{v}_m^2\\ \text{Potential energy: }{V}_1=0\\ \text{Now, applying principles of conservation of energy and simple harmonic motion}\text{.}\\ T_1+V_1=T_2+V_2:\\ \frac{1}{2}{m}_s{v}_m^2+\frac{1}{4}\rho V{v}_m^2+0=0+\frac{1}{2}k{x}_m^2\\ \because \left(v_m={\dot{x}}_m=x_m{\omega }_n\right)\\ \frac{1}{2}\left(m_s+\frac{1}{4}\rho V\right)x_m^2{\omega }_m^2=\frac{1}{2}k{x}_m^2\\ {\omega }_n^2=\frac{k}{m_s+\frac{1}{2}\rho V}\\ substitutingvalues,\\ {\omega }_n^2=\frac{500N\text{/}m}{\left(0.5kg\right)+\left(\frac{1}{2}\rho V\right)}\\ \frac{1}{2}\rho V=\frac{1}{2}\left(1000kg\text{/}{m}^3\right)\left(\frac{4}{3}\pi {\left(0.08m\right)}^3\right)\\ \frac{1}{2}\rho V=1.0723kg\\ {\omega }_n^2=\frac{500N/m}{\left(0.5kg\right)+\left(1.0723kg\right)}=318{\overline{s}}^2\\ \text{Period of vibration}\text{.}\\ {\tau }_n=\frac{2\pi }{{\omega }_n}=\frac{2\pi }{\sqrt{318}}\\ \Rightarrow {\tau }_n=0.352s\end{array}$

Conclusion:

The period of vibration of the shell when it is displaced vertically and released is ${\tau }_n=0.352s$.

To determine

(b)

The period of vibration of the shell if the tank is accelerated upward at the constant rate of 8 m/s².

Answer to Problem 19.98P

The period of vibration: ${\tau }_n=0.352s$

Explanation of Solution

Given information:

Mass of the hollow-spherical shell is 500-g.

Radius of the spherical shell is 80 mm.

Constant of the spring is 500 N/m.

Calculations:

$\begin{array}{l}\text{Accelerating the tank vertically upward does not change the mass}\text{. }\\ \text{Hence the period of vibration remains same as calculated in part }\left(a\right).\end{array}$

For the two positions as shown in the figure:

$\begin{array}{l}\text{At Position 2:}\\ \\ \text{Kinetic energy: }{T}_2=0\\ \text{Potential energy: }{V}_2=\frac{1}{2}k{x}_m^2\end{array}$

$\begin{array}{l}\text{At Position 1 :}\\ \text{Kinetic energy: }{T}_1=T_{spere}+T_{fluid}=\frac{1}{2}{m}_s{v}_m^2+\frac{1}{4}\rho V{v}_m^2\\ \text{Potential energy: }{V}_1=0\\ \text{Now, applying principles of conservation of energy and simple harmonic motion}\text{.}\\ T_1+V_1=T_2+V_2:\\ \frac{1}{2}{m}_s{v}_m^2+\frac{1}{4}\rho V{v}_m^2+0=0+\frac{1}{2}k{x}_m^2\\ \because \left(v_m={\dot{x}}_m=x_m{\omega }_n\right)\\ \frac{1}{2}\left(m_s+\frac{1}{4}\rho V\right)x_m^2{\omega }_m^2=\frac{1}{2}k{x}_m^2\\ {\omega }_n^2=\frac{k}{m_s+\frac{1}{2}\rho V}\\ substitutingvalues,\\ {\omega }_n^2=\frac{500N\text{/}m}{\left(0.5kg\right)+\left(\frac{1}{2}\rho V\right)}\\ \frac{1}{2}\rho V=\frac{1}{2}\left(1000kg\text{/}{m}^3\right)\left(\frac{4}{3}\pi {\left(0.08m\right)}^3\right)\\ \frac{1}{2}\rho V=1.0723kg\\ {\omega }_n^2=\frac{500N/m}{\left(0.5kg\right)+\left(1.0723kg\right)}=318{\overline{s}}^2\\ \text{Period of vibration}\text{.}\\ {\tau }_n=\frac{2\pi }{{\omega }_n}=\frac{2\pi }{\sqrt{318}}\\ \Rightarrow {\tau }_n=0.352s\end{array}$

The period of vibration of the shell when it is displaced vertically and released is ${\tau }_n=0.352s$.