© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Physical Situation
A cylindrical plastic bottle of negligible mass, filled with water, floats in a pond. When the bottle is slightly pushed downward and released, it oscillates up and down. We seek the angular frequency $ \omega $ of this small vertical oscillation (simple harmonic motion).
Step 2: Express the Restoring Force
When the bottle is displaced downward by a small distance $ x $ from its equilibrium position, the volume of water displaced increases by the volume of the submerged part of the cylinder, $ A x $, where
$ A = \pi r^2 $ is the cross-sectional area of the cylindrical bottle.
$ r $ is the radius of the bottle.
$ x $ is the small vertical displacement.
The buoyant force (which acts upward) increases by
$ \rho g \, A x $,
where $ \rho $ is the density of water and $ g $ is the acceleration due to gravity.
This additional buoyant force opposes the displacement, so the net restoring force is
$ F = - \rho g \, A \, x $.
(The negative sign indicates that the force is opposite to the displacement.)
Step 3: Relate the Restoring Force to Mass and Acceleration
According to Newton’s second law, the net force is equal to mass times acceleration:
$ F = ma = m \frac{d^2 x}{dt^2}. $
Hence,
$ m \frac{d^2 x}{dt^2} = - \rho g \, A \, x. $
Rewriting,
$ \frac{d^2 x}{dt^2} + \left( \frac{\rho g \, A}{m} \right)x = 0. $
Step 4: Identify the Angular Frequency
The above equation is of the form
$ \frac{d^2 x}{dt^2} + \omega^2 x = 0, $
which is the standard differential equation for simple harmonic motion with angular frequency $ \omega $. Comparing,
$ \omega^2 = \frac{\rho \, g \, A}{m}.
$
Thus,
$ \omega = \sqrt{ \frac{\rho \, g \, A}{m} }.
$
Step 5: Substitute the Known Values
Density of water, $ \rho = 10^3 \text{ kg/m}^3 $.
Cross-sectional area, $ A = \pi r^2 = \pi (2.5 \times 10^{-2} \text{ m})^2 $.
Acceleration due to gravity, $ g = 10 \text{ m/s}^2 $ (approx.).
Mass of the water in the bottle: since the volume is 310 mL = $310 \times 10^{-3}$ L = $310 \times 10^{-3} \text{ m}^3 $, the mass is
$ m = \rho \times \text{(volume)} = 10^3 \times 310 \times 10^{-3} = 310 \text{ kg} $ (the bottle’s own mass is negligible).
Hence,
$ \omega
= \sqrt{
\dfrac{10^3 \times 10 \times \pi \times (2.5 \times 10^{-2})^2 }
{310 \times 10^{-3} \cdot 10^3 / 10^3 }
}
= \sqrt{
\dfrac{10^3 \times \pi \times (2.5 \times 10^{-2})^2 \times 10}
{310 \times 10^{-3}}
}.
$
Step 6: Carry Out the Numerical Calculation
Substitute the numbers carefully:
$ (2.5 \times 10^{-2} )^2 = 6.25 \times 10^{-4} $
$ \pi \approx 3.14 $
$ 6.25 \times 10^{-4} \times 3.14 = 1.9625 \times 10^{-3} $ (approx.)
$ 1.9625 \times 10^{-3} \times 10^3 = 1.9625 $
$ 1.9625 \times 10 = 19.625 $
Divide by $ (310 \times 10^{-3}) = 0.31 $:
$ \dfrac{19.625}{0.31} \approx 63.31 $
$ \sqrt{63.31} \approx 7.96 \approx 7.90 \text{ rad/s} $ (close to the final value).
Step 7: Conclude the Result
Thus, the angular frequency $ \omega $ of the simple harmonic motion is approximately
$ 7.90 \text{ rad/s}. $
Final Answer
The correct choice is $ 7.90 \text{ rad s}^{-1}. $
