A wooden block of mass 0.5 kg and density 800 `kgm^-3` is fastened to the free end of a vertical spring of spring constant 50 `Nm^-1` 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.

Answer

A, B, C

Explanation

Given m = 5kg, `P=800kg/m^3, k=50N/m`
a. From the free body diagram
`[here Vrho_ggtmg.` So, elongation will be there]
`mg=kX+Vrho_ug`
`rarr (0.5)x(10+50x(X))`
`=(0.5/500)xx10^3xx(10)`
`rarr 50x=(0.5)xx(10)xx((10/8)-1)`
`rarr 50x=(5/4)`
`rarr x=(5/200)`
`=x=2.5cm`
b. since the system is inside water only kx unbalance force will be the driving force for SHM. There is no change in buoyant force
So, `ma=kx`
`rarr a=kx/m`
`rarrw^2x=kx/m`
`rarr ((2p)/T)^2=k/m`
`rarr T=2pisqrt(m/k)`
`=2pixxsqrt((0.5/50))=pi/5sec`