Question

A cylindrical object of outer diameter 10 cm, height 20 cm and density 8000 kg m⁻³ is supported by a vertical spring and is half dipped in water as shown in figure. (a) Find the elongation of the spring in equilibrium condition. (b) If the object is slightly depressed and released, find the time period of resulting oscillations of the object. The spring constant = 500 N m⁻¹.

Answer 1

Given:
Diameter of the cylindrical object, d = 10 cm
∴ Radius, r = 5 cm
Height of the object, h = 20 cm
Density of the object, ρ_b = 8000 kg/m³ = 8 gm/cc
Density of water, ρ_w = 1000 kg/m³
Spring constant, k = 500 N/m = 500 × 10³ dyn/cm
(a) Now, in the equilibrium the weight of the object is balanced by the bouyant force and the spring force.

Therefore,
F + U = mg [F = kx]
Here, U is the upward thrust and x is the small displacement from the equilibrium position.
kx + Vρ_wg = mg
⇒ $500\times {10}^3\times \left(x\right)+\pi r^2\times h\times 10\times 1000=\pi r^2\times h\times {\rho }_b\times 10$

⇒ $500\times {10}^3\times \left(x\right)=\pi r^2\times h({\rho }_b-1000)=\pi \times (5{)}^2\times 20\times (8000-1000)$

⇒ $50x=\pi \times 25\times 2\times \rho b-12x=\pi (8-0.5)or,x=\pi \times 7.5=23.5cm$
(b) We know that X is the displacement of the block from the equilibrium position.
Now,
Driving force:
F=kX+Vρw×g⇒ma=kx+πr2×(X)×ρw×g=(k+πr2×ρw×g)x⇒ω2×(X)=k+πr2×ρw×gm×(X)In SHM a=ω2X, time period T is T=2πmk+πr2×ρw×g =2ππ×25×20×8500+103+π×25×1×1000 =0.935 s