A cylindrical object of outer diameter 10 cm, height 20 cm and de… - Vidyadip

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.

Find the elongation of the spring in equilibrium condition.
If the object is slightly depressed and released, find the time period of resulting oscillations of the object. The spring constant $= 500N/m$.

✓

Answer

The volume of the object $=\pi\times5^2\times20\text{cm}^3=500\pi\text{cm}^3$
The mass of the object $=500\pi\times\frac{8000}{10^6}\text{Kg}$
$=4\pi\text{Kg}$
Since the object is half dipped, the volume of the water displaced
$=\frac{500\pi}{2}$
$=250\pi\text{cm}^3$
The mass of the water displaced $=\frac{250\pi\times1000}{10^6}\text{Kg}$
$=0.25\pi\text{Kg}$
The force of buoyancy $=0.25\pi\text{g}\ \text{N}$
Hence the apparent weight of the object $=(4\pi-0.25\pi)\text{g}\ \text{N}$
$=3.75\pi\text{g}\ \text{N}$
The spring constant $\text{K}=500\text{N}/\text{m}$

Hence the elongation of the spring

$=\frac{3.75\pi\text{g}}{500}\text{m}=3.75\times3.14\times\frac{10}{500}\text{m}$
Taking $g = 10 m/s^2$
$=0.235\text{m}$
$=23.5\text{cm}$

Let the object be dipped to a distance $X m.$ Assuming the axis of the cylindrical object vertical, the extra volume displaced$=\frac{\pi\times5^2\times\text{x}}{10000}\text{m}^3$

The buoyancy force $=\Big(\frac{\pi\times5^2\times\text{x}}{10000}\text{m}^3\Big)\times1000\text{g}\ \text{N}$
$=2.5\pi\text{g}\text{x}\ \text{N}$
The spring force $=\text{KxN}$
Hence the total upward force $=2.5\pi\text{g}\text{x}+500\text{X}\ \text{N}$
Hence the vertical acceleration at the moment $=\frac{\text{Force}}{\text{Mass}}$
$=\frac{\big(2.5\pi\text{x}+500\text{X}\big)}{4\pi}\text{m}/\text{s}^2$
$=\frac{\big(2.5\pi\text{g}+500\big)\text{X}}{4\pi}\text{m}/\text{s}^2$
But also this acceleration $=\omega^2\text{X}\text{m}/\text{s}^2$
Equating, $\omega^2\text{x}=\frac{\big(2.5\pi\text{g}+500\big)\text{X}}{4\pi}$
$=46.04\text{X}$
$\rightarrow\omega=\sqrt{(46.04)}=6.785\text{s}^{-1}$
So, the time period
$=\frac{2\pi}{\omega}$
$=\frac{2\pi}{6.785}\text{s}$
$=0.93\text{s}$

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