Problem 9.53 Challenge

พิจารณา F.B.D. ของวัตถุ

\(F_{B}= ρH^{2}yg\)

\(mg = ρ_{0}H^{3}g\)

จาก \(ΣF_{y}= ma_{y}\)

\(F_{B}- mg = -ma_{y}\)

\(ρH^{2}yg - ρ_{0}H^{3}g = -ρ_{0}H^{3}a_{y}\)

\(a_{y}+\frac{ρg}{ρ_{0}H}y - g = 0\)

\(a_{y}+\frac{ρg}{ρ_{0}H}(y -\frac{ρ_{0}H}{ρ}) = 0\)

\(a_{y}+ \omega _{0}^{2}(y -\frac{ρ_{0}H}{ρ}) = 0; \omega _{0} =\sqrt{\frac{ρg}{ρ_{0}H}}\)

\(a_{y}+ \omega _{0}^{2}(y - y_{0}) = 0; y_{0} =\frac{ρ_{0}H}{ρ}\)

พิจารณา \(\frac{\Delta (y + y_{0})}{\Delta t}=\frac{\Delta y}{\Delta t}\)

\(\frac{\Delta (\frac{\Delta (y + y_{0})}{\Delta t})}{\Delta t}=\frac{\Delta (\frac{\Delta y}{\Delta t})}{\Delta t}\)

\(a_{y + y_{0}}= a_{y}\)

\(\therefore a_{y + y_{0}}+ \omega _{0}^{2}(y + y_{0}) = 0= 0 ; \omega _{0} =\sqrt{\frac{ρg}{ρ_{0}H}}\)

จาก \(ω_{0} = \sqrt{\frac{ρg}{ρ_{0}H}}\)

\(\frac{2\pi}{T}=\sqrt{\frac{ρg}{ρ_{0}H}}\)

\(T = 2\pi \sqrt{\frac{ρ_{0}H}{ρg}}\;\blacksquare\)