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, d = 10 cm

$\Rightarrow$ r = 5 cm

h = 20 cm

${\rho }_b=8000kg/m^3$ = 8 gm/cc

k= 500 N/m = $500\times {10}^3\frac{\text{dyne}}{cm}$

(a) Here, F+U =mg [where F = kx]

$\Rightarrow kx+V{\rho }_{w}g$ = mg

$\Rightarrow 500\times {10}^3\times \left(x\right)+\left(\pi r^2\right)\times \left(\frac{h}{2}\right)\times 1\times 1000$

= $\pi r^2\times h\times {\rho }_b\times 1000$

$\Rightarrow 500\times {10}^3\times \left(x\right)$

= $\pi r^2\times h\times 1000(\rho -\frac{1}{2})$

= $\pi \times (5{)}^2\times 20\times 1000({\rho }_b-\frac{1}{2})$

$\Rightarrow 50x=\pi \times 25\times 2\times ({\rho }_b-\frac{1}{2})$

x= $\pi (8-0.5)$

or x = $\pi \times 7.5$ = 23.5 cm

$\Rightarrow ma=kX+\pi r^2\times {\rho }_w\times g(k+\pi r^2\times {\rho }_w\times g)X$

$\Rightarrow {\omega }^2\left(X\right)=\frac{k+\pi r^2\times {\rho }_w\times g}{m}\times$ (X)

[because a = $w^2$ X in SHM]

$\Rightarrow T=2\pi \frac{m}{k+\pi r^2\times g}$

= $2\pi \frac{\pi \times 25\times 20\times 8}{500+{10}^3+\pi \times 25\times 1\times 1000}$

= 0.935 s.