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.
Answers
As per as the Principle of the Flotation,
Weight of the cylindrical object = Weight of the liquid displaced.
Weight of the cylindrical object = mass × g
= 2 × g
Weight of the Fluid Displaced = Upthrust
= Volume of of solid immersed × density of water × g
= πr²h × 1 × g
where, h is the height and r is the radius of the solid immersed in a water.
∴ Weight of the fluid displaced = π(20)²h × 1 × g
2000 × g = π(10)²h × 1 × g
2000 = π(10)²h
2000 = π × 100 × h
∴ h = 20/π cm.
Now, Using the formula of the Time Period,
T = 2π√(h/g)
T = 2 × 22/7 × √(20/π × 980)
∴ T = 0.51 seconds.
Hence, the time period of the S.H.M. is 0.51 seconds.
Hope it helps.
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 =k²X, equating both we get,
k²X =5πgX
→k=√(5πg)
Hence the time period of the simple harmonic motion =2π/k
=2π/√(5πg) =2*3.14/√(5π*9.8) =0.50 s