A cylindrical block of mass M and area of cross section A is floating in a liquid of density with its axis vertical. When depressed a little and released the block starts oscillating. The period of…

Given: A cylindrical block of mass M and cross-sectional area A floats vertically in a liquid of density. It is depressed by a small distance x and released. Find: The time period of small oscillations. For a small downward displacement x, the extra displaced volume is Ax. Therefore the additional buoyant force acting upward is F = A g x This force is restoring in nature. Using Newton's second law for vertical motion, M d^2xdt^2 = - A g x Rearranging the equation of motion, d^2xdt^2 = - A gMx Comparing with the standard SHM form d^2xdt^2 = - ^2 x we get ^2 = A gM and hence = A gM For small vertical oscillations of a floating body, restoring force is proportional to displacement: F -x. Here the proportionality constant is A g, so the effective spring constant is k = A g. Then T = 2 Mk = 2 M A g Therefore, the correct option is C.

A cylindrical block of mass M and area of cross section A is floa… - Solution | JEE Physics

MCQEasyJEE 2026Simple Harmonic Motion (SHM)

A cylindrical block of mass $M$ and area of cross section $A$ is floating in a liquid of density $\rho$ with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is _____.

A
$2\pi\sqrt{\dfrac{\rho A}{Mg}}$
B
$\pi\sqrt{\dfrac{\rho A}{Mg}}$
C
$2\pi\sqrt{\dfrac{M}{\rho A g}}$
D
$\pi\sqrt{\dfrac{2M}{\rho A g}}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A cylindrical block of mass $M$ and cross-sectional area $A$ floats vertically in a liquid of density $\rho$ . It is depressed by a small distance $x$ and released.

Find: The time period of small oscillations.

For a small downward displacement $x$ , the extra displaced volume is $Ax$ . Therefore the additional buoyant force acting upward is

F = \rho A g x

This force is restoring in nature.

Using Newton's second law for vertical motion,

M\frac{d^2x}{dt^2} = -\rho A g x

Comparison with SHM equation

Rearranging the equation of motion,

\frac{d^2x}{dt^2} = -\frac{\rho A g}{M}x

Comparing with the standard SHM form

\frac{d^2x}{dt^2} = -\omega^2 x

we get

\omega^2 = \frac{\rho A g}{M}

and hence

\omega = \sqrt{\frac{\rho A g}{M}}

Direct SHM result

For small vertical oscillations of a floating body, restoring force is proportional to displacement: $F \propto -x$ . Here the proportionality constant is $\rho A g$ , so the effective spring constant is $k = \rho A g$ .

Then

T = 2\pi\sqrt{\frac{M}{k}} = 2\pi\sqrt{\frac{M}{\rho A g}}

Therefore, the correct option is C.

Common mistakes

Using the full buoyant force instead of the change in buoyant force. Only the additional displaced volume $Ax$ contributes to the restoring force, giving $F = \rho A g x$ .
Missing the negative sign in the equation of motion. The buoyant force acts opposite to the displacement, so the motion equation must be restoring: $M\dfrac{d^2x}{dt^2} = -\rho A g x$ .
Inverting the time-period expression. Since $\omega^2 = \dfrac{\rho A g}{M}$ , the period is $T = 2\pi\sqrt{\dfrac{M}{\rho A g}}$ , not $2\pi\sqrt{\dfrac{\rho A}{Mg}}$ .

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Step-by-step solution

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