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  • The bob of a simple pendulum executes simple harmonic motion in water with a period t while the period of oscillation of the bo

The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t_0

in air. Neglecting frictional force of water and given that the density of the bob is (4/3) ×1000 kg/m3. What relationship between t and t_0 is true            

a) t=t_0

                      

B)            

t=t_0/2

C)            

t=2t_0

                   

D)            

t=4t_0

Answers (1)

best_answer

The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t_0

in air. Neglecting frictional force of water and given that the density of the bob is (4/3) ×1000 kg/m3. What relationship between t and t_0 is true            

A) t=t_0

 B) t=t_0/2

C)  t=2t_0

D)  t=4t_0

Answer: C

\\t_0=2\pi \sqrt\frac{t}{g}\\effective\ weight\ of\ bob\ inside\ water,W'=mg-thrust=V\rho'g\\\Rightarrow V\rho g _{eff}=V(\rho-\rho')g,where\ \rho=Density\ of\ bob\\\Rightarrow g_{eff}=(1-\frac{\rho'}{\rho})g\ and\ \rho'=Density\ of\ water\\\Rightarrow t=2\pi\sqrt\frac{l}{g_{eff}}=2\pi\sqrt\frac{l}{(1-\frac{\rho'}{\rho})g}\\\\\therefore \frac{t}{t_0}=\sqrt\frac{1}{(1-\frac{\rho'}{\rho})}=\sqrt\frac{1}{1-\frac{3}{4}}\\(because \ \rho'=10^3kg/m^3 , \rho=\frac{4}{3}\times10^3kg/m^3)\\\Rightarrow t=2t_0

Posted by

sudhir kumar

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