Cork Oscillation Time Period in Liquids: Density Ratio Calculation
This problem involves a floating, oscillating cylindrical cork and relates its time period of oscillation in two different liquids to their densities.
Equilibrium Condition
For a floating object, the weight (mg) equals the buoyant force. Let the cork have mass m and cross-sectional area A. When floating in a liquid of density $\rho$, submerged to a depth h, the buoyant force is $F_B = \rho A h g$. At equilibrium, $mg = \rho A h g$. This means the mass of the cork is constant: $m = \rho A h$.
Restoring Force and SHM
When the cork is displaced vertically by a small distance x from its equilibrium position, the change in buoyant force provides the restoring force.
- Change in submerged depth = x
- Change in buoyant force $\Delta F_B = \rho A x g$.
- This force acts upwards if depressed downwards, hence it's a restoring force: $F_{restore} = -(\rho A g) x$.
This matches the form $F = -kx$ for Simple Harmonic Motion (SHM), where the effective spring constant is $k = \rho A g$.
Time Period of Oscillation
The time period T of SHM is given by $T = 2\pi \sqrt{\frac{m}{k}}$. Substituting $k = \rho A g$, we get:
$T = 2\pi \sqrt{\frac{m}{\rho A g}}$Since $m = \rho A h$, we can also write $T = 2\pi \sqrt{\frac{\rho A h}{\rho A g}} = 2\pi \sqrt{\frac{h}{g}}$. This shows the time period depends on the equilibrium submerged depth and gravity.
Alternatively, focusing on the spring constant:
$T \propto \frac{1}{\sqrt{k}} \propto \frac{1}{\sqrt{\rho A g}}$Since m, A, and g are constant for the cork and setup:
$T \propto \frac{1}{\sqrt{\rho}}$Calculating Density Ratio
Let $T_1$ be the time period in liquid 1 (density $\rho_1$) and $T_2$ be the time period in liquid 2 (density $\rho_2$).
- Given $T_1 = T$.
- Given $T_2 = 2T$.
Using the proportionality $T \propto 1/\sqrt{\rho}$:
$\frac{T_1}{T_2} = \frac{1/\sqrt{\rho_1}}{1/\sqrt{\rho_2}} = \sqrt{\frac{\rho_2}{\rho_1}}$Substitute the given time periods:
$\frac{T}{2T} = \sqrt{\frac{\rho_2}{\rho_1}}$ $\frac{1}{2} = \sqrt{\frac{\rho_2}{\rho_1}}$Squaring both sides to find the ratio:
$\left(\frac{1}{2}\right)^2 = \frac{\rho_2}{\rho_1}$ $\frac{1}{4} = \frac{\rho_2}{\rho_1}$Therefore, the value of $\rho_2/\rho_1$ is $\frac{1}{4}$.
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$x_1 = \sqrt{7} \sin 5t \text{ cm}$ and $x_2 = 2\sqrt{7} \sin \left(5t + \frac{\pi}{3}\right) \text{ cm}$
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