A cylindrical plastic bottle of negligible mass is filled with 310 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 cm then $$\omega$$ is close to: (density of water $$= 10^3$$ kg/m$$^3$$)

We have a cylindrical plastic bottle whose own mass is negligible but which is filled with water of volume 310 ml. Remember that

$$1\;\text{ml}=1\;\text{cm}^3=1\times10^{-6}\;\text{m}^3,$$

so the volume of water inside the bottle is

$$V = 310\;\text{ml}=310\times10^{-6}\;\text{m}^3 = 3.10\times10^{-4}\;\text{m}^3.$$

The density of water is given as $$\rho = 10^{3}\;\text{kg\,m}^{-3},$$ therefore the mass of the water (and hence of the whole bottle-water system, because the bottle itself is negligibly light) is

$$m = \rho V = (10^{3}\;\text{kg\,m}^{-3})(3.10\times10^{-4}\;\text{m}^3)=0.310\;\text{kg}.$$

The bottle floats upright in the pond. When it is pushed slightly downward through a small vertical distance $$x$$ and then released, the only net restoring force is the extra buoyant force that arises because an extra volume of water has been displaced.

By Archimedes’ principle, the upward buoyant force equals the weight of the displaced water. If the cross-sectional area of the bottle is $$A,$$ pushing the bottle downward by $$x$$ submerges an additional volume $$A x$$ of water. The mass of this extra displaced water is $$\rho A x,$$ so its weight (and hence the restoring force) is

$$F_{\text{buoyant}} = \rho g (A x).$$

This force acts upward, opposite to the downward displacement, so the net restoring force is

$$F = -\,\rho g A\,x.$$

This is of the same mathematical form as Hooke’s law $$F=-k x$$ with an effective spring constant

$$k = \rho g A.$$

The equation of motion for the vertical oscillation is therefore

$$m\,\frac{d^{2}x}{dt^{2}} + \rho g A\,x = 0.$$

The standard form of simple harmonic motion is $$m\,\ddot x + k x = 0,$$ whose angular frequency is

$$\omega = \sqrt{\frac{k}{m}}.$$

Substituting $$k=\rho g A$$ gives

$$\omega = \sqrt{\frac{\rho g A}{m}}.$$

Now we compute the cross-sectional area. The radius of the bottle is

$$r = 2.5\;\text{cm} = 0.025\;\text{m},$$

so

$$A = \pi r^{2} = \pi (0.025\;\text{m})^{2} = \pi (6.25\times10^{-4}\;\text{m}^{2}) = 1.9635\times10^{-3}\;\text{m}^{2}.$$

Next, calculate the effective spring constant:

$$k = \rho g A = (10^{3}\;\text{kg\,m}^{-3})(9.8\;\text{m\,s}^{-2})(1.9635\times10^{-3}\;\text{m}^{2}).$$

First multiply $$9.8$$ by $$1.9635\times10^{-3},$$ obtaining

$$9.8 \times 1.9635\times10^{-3} = 1.9246\times10^{-2}.$$

Then multiply by $$10^{3}$$:

$$k = 10^{3}\times1.9246\times10^{-2}\;\text{N\,m}^{-1} = 19.246\;\text{N\,m}^{-1}.$$

Finally, substitute $$k$$ and $$m=0.310\;\text{kg}$$ into the formula for $$\omega$$:

$$\omega = \sqrt{\frac{19.246\;\text{N\,m}^{-1}}{0.310\;\text{kg}}} = \sqrt{62.086\;\text{s}^{-2}} = 7.88\;\text{rad\,s}^{-1}.$$

The computed value $$7.88\;\text{rad\,s}^{-1}$$ is very close to $$7.9\;\text{rad\,s}^{-1}$$ listed among the options.

Hence, the correct answer is Option C.