A wooden block of mass 0.5 kg and density 800 kg/m^3 is fastened to the free end of a vertical spring of spring constant 50 N/m
A wooden block of mass 0.5 kg and density 800 kg/m3 is fastened to the free end of a vertical spring of spring constant 50 N/m fixed at the bottom. If the entire system is completely immersed in water, find (a) the elongation (or compression) of the spring in equilibrium and (b) the time-period of vertical oscillations of the block when it is slightly depressed and released.
1 Answers
(a) The volume of wooden block = 0.5/800 m³
=0.000625 m³
The force of buoyancy =0.000625*1000 kg =0.625 kg
The Net force on the block in the water =0.625-0.500 kg-weight
=0.125*10 N {Taking g = 10 m/s²}
=1.25 N upward.
Spring constant k=50 N/m
Hence the elongation of the spring =(1.25/50)*100 cm
=2.5 cm
(b) Let the block be depressed by X cm.
The force on the block by the spring =50X N
Hence the acceleration =50X/0.5 =100X m/s²
So, ⍵²X =100X
→⍵ = 10 s⁻¹
Hence time period T = 2π/⍵
=2π/10 s
= π/5 s