A bob of simple pendulum executes shm in water

Asked by admin @ 20/08/2021 in Physics viewed by 277 People

The bob of a simple pendulum executes SHM in water with time period T, while its period of oscillation is T' in air. Given that density of the bob is 4000/3 kg m^(-3) . Find the relation between T andT' by equating torque.

Answered by admin @ 20/08/2021

Let   V be the volume of the bob.Thus mass of the bob   M=ρbV=43×1000VkgAcceleration due to gravity in air is   g.Effective weight of the bob in water   Mg′=Mg−B   where    buoyant force   B=(Vρwater)g43×1000Vg′=43×1000Vg−V×1000g   (as ρwater=1000kg/m3)
43×1000Vg′=13×1000Vg  ⟹g′=g4
Now time period of the bob in air   to=2π√lgTime period of the bob in water t=2π√lg′  =2π√4lg    =2×2π√lg
⟹t=2to

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A bob of simple pendulum executes shm in water

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