Question

A cylindrical body of mass m and cross section A is floating in a liquid of density $\rho_{L}$ such that its axis is vertical. If body is displaced by a small displacement 'x' vertically, find the time period of oscillation of the body :

• A. $2\pi \sqrt{\frac{m}{\rho_{L}Ag}}$
• B. $3\pi \sqrt{\frac{m}{\rho_{L}Ag}}$
• C. $4\pi \sqrt{\frac{m}{\rho_{L}Ag}}$
• D. $5\pi \sqrt{\frac{m}{\rho_{L}Ag}}$

Hint:

For any floating body undergoing vertical oscillations, the restoring force constant 'k' is always equal to the product of the liquid density, the cross-sectional area at the waterline, and the acceleration due to gravity ($k = \rho A g$). You can use this standard result directly to save time in the exam.

Answer 1

The Correct Option is A

Step 1: Physical idea used

When a floating cylinder is pushed down slightly and released, its center of mass moves vertically. Due to gravity and buoyancy, the system behaves like an oscillating system.

Instead of using force comparison, we use the energy method to identify Simple Harmonic Motion.

---

Step 2: Change in buoyant force on displacement

Let the cylinder be pushed down by a small distance x.

This increases the submerged volume by:

Extra volume = area multiplied by x

Increase in buoyant force = density of liquid multiplied by g multiplied by area multiplied by x

---

Step 3: Change in potential energy

Work done by this extra buoyant force is stored as potential energy.

Potential energy gained = one half multiplied by density of liquid multiplied by g multiplied by area multiplied by x square

This expression is of the form:

Potential energy = one half multiplied by k multiplied by x square

Hence, the effective spring constant is:

k = density of liquid multiplied by area multiplied by g

---

Step 4: Time period of oscillation

For a system executing simple harmonic motion, the time period is:

Time period = 2 multiplied by pi multiplied by square root of (mass divided by spring constant)

Substituting the value of k:

Time period = 2 multiplied by pi multiplied by square root of (mass divided by density of liquid multiplied by area multiplied by g)

---

Final Answer:

The time period of oscillation of the floating cylinder is
2 pi multiplied by square root of (m divided by density of liquid multiplied by area multiplied by g)