Chapter 5, Problem 5.19P

To determine

Relationship in dimensionless form and result if Y is negligible.

Answer to Problem 5.19P

The relationship in dimensionless form is $T{\left(\frac{g}{\lambda }\right)}^{1/2}=f_{cn}\left(\frac{h}{\lambda } \right)$.

The period of oscillation T, density ρ, wavelength λ, depth h, gravity g, and surface tension Y the variables and their dimensions.

$T=f_{cn}(\rho ,\lambda ,h,g,Y)$

$\left(T\right)= \left(\frac{M}{L^3}\right)\left(L\right)\left(L\right) \left(\frac{L}{{\text{T}}^2}\right)\left(\frac{M}{{\text{T}}^2}\right)$

Then n = 6 and j = 3, hence we expect n - j = 6 - 3 = 3

Capable of various arrangements and selected by self follows:

$T=f_{cn}(\rho ,\lambda ,h,g,Y)$

$\left(T\right)= \left(\frac{M}{L^3}\right)\left(L\right)\left(L\right) \left(\frac{L}{{\text{T}}^2}\right)\left(\frac{M}{{\text{T}}^2}\right)$

$computing\mathrm{var}iables,$

$\left(T\right)= \left(\frac{M}{L^3}\right)\left(L\right)\left(L\right) \left(\frac{L}{{\text{T}}^2}\right)\left(\frac{M}{{\text{T}}^2}\right)$

$weget,$

${\pi }_1={\left[MLT\right]}^0={\left(\frac{M}{L^3}\right)}^a{\left(\frac{L}{{\text{T}}^2}\right)}^b{\left(L\right)}^c\gamma$

${\pi }_1={\left[MLT\right]}^0={\left(\frac{M}{L^3}\right)}^a{\left(\frac{L}{{\text{T}}^2}\right)}^b{\left(L\right)}^c\left(\frac{M}{{\text{T}}^2}\right)$

$Wehave,$

$a+1=0,-3a+b+c=0,-2c-2=0$

$On\text{}solving,$

$a=-1,b=-2,c=-2$

${\pi }_1=\frac{\gamma }{\rho g{h}^2}$

${\pi }_2={\left[MLT\right]}^0={\left(\frac{M}{L^3}\right)}^e{\left(L\right)}^f{\left(\frac{L}{{\text{T}}^2}\right)}^g\lambda$

$onsolving,$

$\begin{array}{l}e=0,f=-1,g=0\\ {\pi }_2=\frac{\lambda }{h}\end{array}$

${\pi }_3={\left[MLT\right]}^0={\left(\frac{M}{L^3}\right)}^x{\left(L\right)}^y{\left(\frac{L}{{\text{T}}^2}\right)}^z\left(T\right)$

$\begin{array}{l}Onsolving,\\ x=0,y=\frac{1}{2},z=-\frac{1}{2}\end{array}$

${\pi }_3=\left(T\right){\left(\frac{g}{h}\right)}^{\frac{1}{2}}$

Replacing h by $\lambda$,

We get,

Typical result = $T{\left(\frac{g}{\lambda }\right)}^{1/2}=f_{cn}\left(\frac{h}{\lambda } , \frac{Y}{\rho g{\lambda }^2}\right)$

If Y is negligible, ρ drops out also, leaving $T{\left(\frac{g}{\lambda }\right)}^{1/2}=f_{cn}\left(\frac{h}{\lambda } \right)$.

Explanation of Solution

The period of oscillation T, density ρ, wavelength λ, depth h, gravity g, and surface tension Y the variables and their dimensions.

$T=f_{cn}(\rho ,\lambda ,h,g,Y)$

$\left(T\right)= \left(\frac{M}{L^3}\right)\left(L\right)\left(L\right) \left(\frac{L}{{\text{T}}^2}\right)\left(\frac{M}{{\text{T}}^2}\right)$

Then n = 6 and j = 3, hence we expect n - j = 6 - 3 = 3

Capable of various arrangements and selected by self follows:

$T=f_{cn}(\rho ,\lambda ,h,g,Y)$

$\left(T\right)= \left(\frac{M}{L^3}\right)\left(L\right)\left(L\right) \left(\frac{L}{{\text{T}}^2}\right)\left(\frac{M}{{\text{T}}^2}\right)$

$computing\mathrm{var}iables,$

$\left(T\right)= \left(\frac{M}{L^3}\right)\left(L\right)\left(L\right) \left(\frac{L}{{\text{T}}^2}\right)\left(\frac{M}{{\text{T}}^2}\right)$

$weget,$

${\pi }_1={\left[MLT\right]}^0={\left(\frac{M}{L^3}\right)}^a{\left(\frac{L}{{\text{T}}^2}\right)}^b{\left(L\right)}^c\gamma$

${\pi }_1={\left[MLT\right]}^0={\left(\frac{M}{L^3}\right)}^a{\left(\frac{L}{{\text{T}}^2}\right)}^b{\left(L\right)}^c\left(\frac{M}{{\text{T}}^2}\right)$

$Wehave,$

$a+1=0,-3a+b+c=0,-2c-2=0$

$On\text{}solving,$

$a=-1,b=-2,c=-2$

${\pi }_1=\frac{\gamma }{\rho g{h}^2}$

${\pi }_2={\left[MLT\right]}^0={\left(\frac{M}{L^3}\right)}^e{\left(L\right)}^f{\left(\frac{L}{{\text{T}}^2}\right)}^g\lambda$

$onsolving,$

$\begin{array}{l}e=0,f=-1,g=0\\ {\pi }_2=\frac{\lambda }{h}\end{array}$

${\pi }_3={\left[MLT\right]}^0={\left(\frac{M}{L^3}\right)}^x{\left(L\right)}^y{\left(\frac{L}{{\text{T}}^2}\right)}^z\left(T\right)$

$\begin{array}{l}Onsolving,\\ x=0,y=\frac{1}{2},z=-\frac{1}{2}\end{array}$

${\pi }_3=\left(T\right){\left(\frac{g}{h}\right)}^{\frac{1}{2}}$

Replacing h by $\lambda$,

We get,

Typical result = $T{\left(\frac{g}{\lambda }\right)}^{1/2}=f_{cn}\left(\frac{h}{\lambda } , \frac{Y}{\rho g{\lambda }^2}\right)$

If Y is negligible, ρ drops out also, leaving $T{\left(\frac{g}{\lambda }\right)}^{1/2}=f_{cn}\left(\frac{h}{\lambda } \right)$.