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Question Number 60121 by rahul 19 last updated on 18/May/19

Commented by rahul 19 last updated on 18/May/19

$$timeperiodisgivenas:2\pi \sqrt{\frac{\rho L}{\sigma g}}.$$$${Isn}^{\prime }tdatainsufficient?$$

Commented by mr W last updated on 18/May/19

$$dataisenough.$$$$answeraiscorrect.$$

Commented by mr W last updated on 18/May/19

$$foranobjectwithuniformcrosssection,$$$$todeterminethefrequencyortime$$$$periodofitsverticaloscillationinaliquid,$$$$we\mathit{o}\mathit{n}\mathit{l}\mathit{y}needtoknowhowmuchof$$$$itisimmersedwhenitisfloatingin$$$$theliquid.$$$$f=\frac{1}{2\pi }\sqrt{\frac{g}{L_w}}$$$$T=2\pi \sqrt{\frac{L_w}{g}}$$$$with{L}_w=depthimmersedinliquid$$

Answered by mr W last updated on 18/May/19

$$letlengthofpole=L,crosssection=A$$$$lengthimmersedinliquidwhenfloating=L_w$$$$densityofpole=\rho$$$$densityofliquid={\rho }_w$$$$whenfloating\Rightarrow A{\rho }_w{gL}_w=A\rho gL\Rightarrow {\rho }_w{L}_w=\rho L$$$$T=2\pi \sqrt{\frac{L\rho }{{\rho }_{w}g}}=2\pi \sqrt{\frac{{\rho }_w{L}_w}{{\rho }_{w}g}}=2\pi \sqrt{\frac{L_w}{g}}$$$$=2\pi \sqrt{\frac{0.8}{9.81}}=\frac{3.998\pi }{7}\approx \frac{4\pi }{7}$$$$\Rightarrow answera$$

Commented by rahul 19 last updated on 19/May/19

$$thankyousir.$$

Answered by tanmay last updated on 18/May/19

$$wtofpole=upthrust$$$$mg=\pi R^{2}x{\rho }_{liquid}g$$$$\pi R^{2}h{\rho }_{pole}g=\pi R^{2}x{\rho }_{liwuid}g$$$$h=lengthofpole$$$$x=submergedlengthofpole$$$$whenpushedfurtherby△x$$$$thenupthrust>wtofpole$$$$sorestoringforce$$$$=\pi R^2(x+△x){\rho }_{liq}g-\pi R^{2}h{\rho }_{pole}g$$$$=\pi R^{2}x{\rho }_{liq}g-\pi R^{2}h{\rho }_{pole}g+\pi R^2.△x.{\rho }_{liq}g$$$$=\pi R^2.△x.{\rho }_{liq}g$$$$accelaration(w^2△x)$$$$=\frac{\pi R^2.△x.{\rho }_{liq}g}{\pi R^{2}h{\rho }_{pole}}$$$$=\frac{{\rho }_{liq}\times g}{{\rho }_{pole}\times h}\times △x$$$$so{w}^2△x=\frac{{\rho }_{liq}g}{{\rho }_{pole}h}\times △x$$$$\frac{2\pi }{T}=\sqrt{\frac{{\rho }_{liq}g}{{\rho }_{pole}h}}$$$$T=2\pi \times \sqrt{\frac{{\rho }_{pole}}{{\rho }_{liq}}\times \frac{h}{g}}$$$$inquestionnomentionofvalueof$$$$densitypole\left({\rho }_{pole}\right)$$$$densityliquid({\rho }_{liquid)}$$$$lengthofpole\left(h\right)$$$$now..fromfirsteqn$$$$mg=\pi R^{2}x{\rho }_{liq}g$$$$\pi R^{2}h{\rho }_{pole}g=\pi R^{2}x{\rho }_{liq}g$$$$h{\rho }_{pole}=x{\rho }_{liquid}$$$$\frac{h{\rho }_{pole}}{{\rho }_{liq}}=x$$$$T=2\pi \sqrt{\frac{x}{g}}$$$$givenx=80cm$$$$$$

Commented by rahul 19 last updated on 19/May/19

$$thankusir.$$

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