Physics - SHM Question with Solution | TestHub

Case-I

In equilibrium

${\mathrm{F}}_{\mathrm{up}}=\mathrm{mg}$

$V_i{\rho }_{L}g=mg$

$Ah{\rho }_{L}g=AL{\rho }_{s}g\dots \dots \dots \left(1\right)$



Case-II

Let block is pressed by x more in liquid

$F_{net}=F_{up}^1-mg$

$={\mathrm{V}}_{\mathrm{i}}^1{\mathrm{\rho }}_{\mathrm{L}}\mathrm{g}-\mathrm{mg}$

${\mathrm{F}}_{\mathrm{net}}=\mathrm{A}\left(\mathrm{h}+\mathrm{x}\right){\mathrm{\rho }}_{\mathrm{L}}\mathrm{g}-\mathrm{AL}{\mathrm{\rho }}_{\mathrm{s}}\mathrm{g}$

${\mathrm{F}}_{\mathrm{net}}=\mathrm{Ah}{\mathrm{\rho }}_{\mathrm{L}}\mathrm{g}+\mathrm{Ax}{\mathrm{\rho }}_{\mathrm{L}}\mathrm{g}-\mathrm{AL}{\mathrm{\rho }}_{\mathrm{s}}\mathrm{g}$. . . . . . . . . . (2)

from equation (1) & (2)

${\mathrm{F}}_{\mathrm{net}}=\mathrm{Ax}{\mathrm{\rho }}_{\mathrm{L}}\mathrm{g }$restoring force in upward direction

$\alpha =\frac{F_{net}}{m}=\left(\frac{A{\rho }_{L}g}{A{\rho }_{s}L}\right)x$

$\mathrm{\alpha }={\mathrm{\omega }}^2\mathrm{x\; for\; SHM}$

and${\mathrm{\omega }}^2=\frac{\mathrm{g}}{\mathrm{l}}$

so$\frac{\mathrm{g}}{\mathrm{l}}=\frac{{\mathrm{\rho }}_{\mathrm{L}}\mathrm{g}}{{\mathrm{\rho }}_{\mathrm{s}}\mathrm{L}}$

so$\mathrm{l}=\frac{{\mathrm{\rho }}_{\mathrm{s}}\mathrm{L}}{{\mathrm{\rho }}_{\mathrm{L}}}=\frac{650\times 54}{900}=39\mathrm{cm}$