Animated Solution for Physics - Oscillations: A cylindrical plastic bottle of negligible mass is filled with $310\text{ mL}$ of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $\omega$. If the radius of the bottle is $2.5\text{ cm}$, then $\omega$ is close to (Take, density of water $=10^3{\text{ kg/m}}^3$)

The problem of a floating object bobbing up and down is a classic in physics, beautifully bridging the worlds of fluid mechanics and simple harmonic motion. Imagine a plastic bottle, filled with water, peacefully floating in a pond. When we disturb this peace by pushing it down, it doesn't just sink or pop out; it dances. Let's decode the physics behind this rhythmic dance.

Before we push the bottle, it is in a state of perfect equilibrium. The downward pull of gravity is exactly balanced by the upward buoyant force from the pond.

Since the problem states the bottle itself has negligible mass, the entire mass of our oscillating system comes from the water inside it. Let's say the bottle sinks to a depth `$l$` in the pond. The volume of the water inside the bottle is simply the cross-sectional area `$A$` multiplied by this depth `$l$`. Therefore, the mass `$m$` of our system is:

`$m=\rho Al$`

where `$\rho$` is the density of water. This equilibrium depth `$l$` is the anchor point for our entire oscillation.

Now, let's introduce a disturbance. We push the bottle down by a small extra distance `$x$`. By doing this, we force the bottle to displace an additional volume of pond water equal to `$Ax$`.

According to Archimedes' principle, this extra displaced water fights back! It creates an upward restoring force equal to the weight of the extra displaced liquid. Since we pushed it down (let's call downward positive), the force acts upwards (negative).

`$F_{\text{res}}=-(\rho Ax)g$`

This is the force that drives the simple harmonic motion. Now, we bring in the heavy hitter: Newton's Second Law of Motion, `$F=ma$`. Equating our restoring force to mass times acceleration, we get:

`$ma=-\rho Agx$`

Remember our expression for the mass of the system? Let's substitute `$m=\rho Al$` into this equation:

`$(\rho Al)a=-\rho Agx$`

Take a moment to appreciate what happens next. The density of water `$\rho$` and the cross-sectional area `$A$` appear on both sides of the equation. They completely cancel out!

`$la=-gx$`

`$a=-\left(\frac{g}{l}\right)x$`

This is a profound result. It tells us that the oscillation doesn't care about how wide the bottle is, or even what liquid it's floating in, as long as the liquid inside the bottle is the same as the liquid outside.

We recognize this equation immediately. It is the defining differential equation of Simple Harmonic Motion, `$a=-{\omega }^{2}x$`. By comparing the two, we find the square of the angular frequency:

`${\omega }^2=\frac{g}{l}$`

`$\omega =\sqrt{\frac{g}{l}}$`

The angular frequency behaves exactly like a simple pendulum of length `$l$`!

To find `$\omega$`, we just need to calculate the equilibrium submerged length `$l$`. We are given the volume `$V=310\text{ mL}$` and the radius `$r=2.5\text{ cm}$`. We must be meticulous with our units, converting everything to meters.

`$V=310\times 10^{-6}{\text{ m}}^3$` `$r=2.5\times 10^{-2}\text{ m}$`

The length `$l$` is the volume divided by the cross-sectional area:

`$l=\frac{V}{\pi r^2}=\frac{310\times 10^{-6}}{\pi (2.5\times 10^{-2}{)}^2}$`

`$l=\frac{310\times 10^{-6}}{\pi \times 6.25\times 10^{-4}}\approx 0.158\text{ m}$`

Now, we substitute this back into our angular frequency formula. Taking `$g=10{\text{ m/s}}^2$`:

`$\omega =\sqrt{\frac{10}{0.158}}\approx \sqrt{63.29}\approx 7.96\text{ rad/s}$`

The angular frequency is approximately `$8\text{ rad/s}$`. Interestingly, if you look at the options provided in the original JEE question, none of them match this correct value. This happens occasionally in competitive exams. The key is to trust your derivation and your physics intuition. The math doesn't lie!