Discussion Forum

The bob of a simple pendulum  performs SHM with period  T in air and with period T₁ in water, Relation  between T and T₁ is (neglect friction due to water , density of the material  of the bob  is = $\frac{9}{8}\times 10^{3}kg m^{3}$ , density of water =1 gcc^-1)

Answer:

Explanation:

 Time period  of simple pendulum in water is given by

                   $T=2\pi\sqrt{\frac{l}{g_{eff}}}$

[ Symbols have their  usual meanings and   $g_{eff}$ = acceleration due to gravity in water]

 As we know, $g_{eff}$$=g\left(\frac{\sigma-\rho}{\sigma}\right)$

 [ where, $\sigma$ = density of bob.$\rho$ = density of water]

 $=9.8\left(\frac{\frac{9}{8}\times10^{3}-10^{3}}{\frac{9}{8}\times 10^{3}}\right)$

   $=9.8\left(\frac{\frac{9}{8}-1}{\frac{9}{8}}\right)=9.8\left(\frac{1}{8} \times \frac{8}{9}\right)$

 =$\frac{9.8}{9}$

  So, $T_{1}=2\pi \sqrt{\frac{l \times 9}{9.8}}\Rightarrow T_{1}=3T$

 [$\because $  Time period of simple pendulum in air, 

                $T_{}=2\pi \sqrt{\frac{l }{9.8}}$]