A cylindrical object of outer diameter 20 cm and mass 2 kg floats in water with its axis vertical. If it is slightly depressed
A cylindrical object of outer diameter 20 cm and mass 2 kg floats in water with its axis vertical. If it is slightly depressed and then released, find the time period of the resulting simple harmonic motion of the object.
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Cross-sectional area of the cylinder =π(0.10)² m²
If the cylinder is depressed to a depth of X meter, the net force on the cylinder upward =πX(0.10)² * 1000 kg-weight.
=10πX kg-weight =10πgX Newton
Mass of the cylinder =2 kg
Acceleration =Force/mass =10πgX/2 =5πgX m/s²
But acceleration is also =⍵²X,
equating both we get, ⍵²X =5πgX
→⍵ =√(5πg)
Hence the time period of the simple harmonic motion
=2π/⍵ =2π/√(5πg)
=2*3.14/√(5π*9.8)
=0.50 s
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