Question

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⁻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 1

Given:
Mass of the wooden block, m = 5 kg
Density of the block, ρ = 800 kg/m³
Spring constant, k = 50 N/m
Density of water, ρ_w = 1000 kg/m³
(a) Using the free body diagram, we get:
mg = kx + Vρ_wg (Here, Vρ_wg > mg. So, there will be some elongation.]
$$\begin{gathered} \Rightarrow (0.5)\times (10+50\times (x\left)\right)=\left(\frac{0.5}{800}\right)\times {10}^3\times \left(10\right) \\ \Rightarrow 50x=\left(0.5\right)\times \left(10\right)\times \left(\left(\frac{10}{8}\right)-1\right) \\ \Rightarrow 50x=\left(\frac{5}{4}\right) \\ \Rightarrow x=2.5\mathrm{cm} \end{gathered}$$

(b) As the system is inside the water, the unbalance force will be the driving force, which is kx for SHM.
Hence, there will be no change in the buoyant force.
$$\begin{gathered} \therefore ma=kx \\ \Rightarrow a=kx/m \\ \Rightarrow w^{2}x=kx/m \\ \Rightarrow {\left(\frac{2p}{\mathrm{T}}\right)}^2=k/m \\ \Rightarrow T=2\mathrm{\pi }\sqrt{\frac{m}{k}} \\ =2\mathrm{\pi }\times \sqrt{\left(\frac{0.5}{50}\right)}=\frac{\mathrm{\pi }}{5}\mathrm{s} \end{gathered}$$