Question

A cone with half the density of water is floating in water as shown in figure. It is depressed down by a small distance \(\delta \, (\delta \ll H)\) and released. The frequency of simple harmonic oscillations of the cone is

• A. \(\dfrac{1}{2\pi}\sqrt{\dfrac{6g}{H}\dfrac{1}{4^{1/3}}}\)
• B. \(\dfrac{1}{2\pi}\sqrt{\dfrac{3g}{H}\dfrac{1}{4^{1/3}}}\)
• C. \(\dfrac{1}{2\pi}\sqrt{\dfrac{6g}{2H}}\)
• D. \(\dfrac{1}{2\pi}\sqrt{\dfrac{g}{H}}\)

Hint:

For floating-body SHM, the restoring force due to small vertical displacement is equal to the extra buoyant force: \[ F=\rho g A\delta \] where \(A\) is the area of cross-section at the waterline.

Answer 1

The Correct Option is A

Step 1: Use the condition of floating equilibrium.
Density of cone is half the density of water.
So, in equilibrium, weight of cone equals weight of displaced water.
Therefore,
\[ \frac{V_{\text{submerged}}}{V_{\text{cone}}}=\frac{1}{2} \]
Let the submerged height from the vertex be \(x\).
Since similar cones have volumes proportional to cube of their heights,
\[ \frac{x^3}{H^3}=\frac{1}{2} \] \[ x=\frac{H}{2^{1/3}} \]

Step 2: Find the area of waterline cross-section.
The semi-vertical angle of the cone is \(30^\circ\).
So, radius at submerged height \(x\) is
\[ r=x\tan 30^\circ \] \[ r=\frac{x}{\sqrt{3}} \] Hence, area of waterline cross-section is
\[ A=\pi r^2 \] \[ A=\pi \frac{x^2}{3} \] Substituting \(x=\dfrac{H}{2^{1/3}}\),
\[ A=\frac{\pi H^2}{3\cdot 2^{2/3}} \]

Step 3: Find the restoring force for small displacement.
When the cone is depressed by a small distance \(\delta\), extra volume of water displaced is approximately
\[ A\delta \] Extra upward buoyant force is
\[ F=\rho_w g A\delta \] This force acts opposite to displacement, so it behaves like restoring force.
Thus, effective spring constant is
\[ k=\rho_w g A \]

Step 4: Find the mass of the cone.
Volume of complete cone is
\[ V=\frac{1}{3}\pi R^2H \] Since \(\tan 30^\circ=\dfrac{R}{H}\),
\[ R=\frac{H}{\sqrt{3}} \] Therefore,
\[ V=\frac{1}{3}\pi \left(\frac{H}{\sqrt{3}}\right)^2H \] \[ V=\frac{\pi H^3}{9} \] Density of cone is \(\dfrac{\rho_w}{2}\).
So, mass of cone is
\[ m=\frac{\rho_w}{2}\cdot \frac{\pi H^3}{9} \] \[ m=\frac{\rho_w \pi H^3}{18} \]

Step 5: Find angular frequency.
For SHM,
\[ \omega^2=\frac{k}{m} \] \[ \omega^2=\frac{\rho_w g A}{m} \] Substituting values,
\[ \omega^2= \frac{\rho_w g \left(\dfrac{\pi H^2}{3\cdot 2^{2/3}}\right)} {\dfrac{\rho_w \pi H^3}{18}} \] \[ \omega^2=\frac{6g}{H\cdot 2^{2/3}} \] Since,
\[ 2^{2/3}=4^{1/3} \] we get
\[ \omega^2=\frac{6g}{H}\cdot \frac{1}{4^{1/3}} \]

Step 6: Find frequency.
\[ f=\frac{\omega}{2\pi} \] \[ f=\frac{1}{2\pi}\sqrt{\frac{6g}{H}\cdot \frac{1}{4^{1/3}}} \]

Step 7: Final conclusion.
Hence, the frequency of simple harmonic oscillations is
\[ \boxed{\frac{1}{2\pi}\sqrt{\frac{6g}{H}\frac{1}{4^{1/3}}}} \]