A cylindrical block of mass M and cross-sectional area A floats in a liquid (density ρ). When displaced vertically and released, it oscillates. We need to find the oscillation period.
At equilibrium, the block's weight (Mg) equals the buoyant force. Let the submerged depth be h. The buoyant force is $ F_B = A h \rho g $. Equating forces:
$ M g = A h \rho g $
If the block is depressed by a small distance x, the submerged depth increases to (h + x). The new buoyant force is $ F_B' = A(h+x)\rho g $. The net upward force, acting as the restoring force, is the difference between the new buoyant force and the weight:
$ F_{restore} = F_B' - Mg $ $ F_{restore} = A(h+x)\rho g - A h \rho g $ $ F_{restore} = A x \rho g $ This force opposes the displacement x, so the restoring force is $ F_{restore} = - A x \rho g $.
Using Newton's second law ($ F=Ma $, where $ a = \frac{d^2x}{dt^2} $):
$ M a = - A x \rho g $ $ M \frac{d^2x}{dt^2} = - A x \rho g $ The equation for SHM is $ \frac{d^2x}{dt^2} = -\omega^2 x $. Comparing, the square of the angular frequency ($ \omega^2 $) is:
$ \omega^2 = \frac{A \rho g}{M} $
The angular frequency is $ \omega = \sqrt{\frac{A \rho g}{M}} $. The period (T) is related to angular frequency by $ T = \frac{2\pi}{\omega} $.
$ T = \frac{2\pi}{\sqrt{\frac{A \rho g}{M}}} $ $ T = 2\pi \sqrt{\frac{M}{A \rho g}} $
The derived period of oscillation is $ T = 2\pi \sqrt{\frac{M}{\rho A g}} $. This expression matches Option 3. The provided correct answer is Option A.
A particle is subjected to two simple harmonic motions as :
$x_1 = \sqrt{7} \sin 5t \text{ cm}$ and $x_2 = 2\sqrt{7} \sin \left(5t + \frac{\pi}{3}\right) \text{ cm}$
where x is displacement and t is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2} \text{ ms}^{-2}$. The value of x is :

A piston of mass $M$ is hung from a massless spring whose restoring force law goes as $F= -kx^3$, where $k$ is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with 'n' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature $T$) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height $L_0$ to $L_1$, the total energy delivered by the filament is : (Assume spring to be in its natural length before heating)
Two simple harmonic motions, as shown below, are at right angles. They are combined to form lissajous figures.
$x(t) = A \sin (at + \delta)$
$y(t) = B \sin (bt)$
Identify the correct match below :
The motion of a mass on a spring, with spring constant K is as shown in figure.
The equation of motion is given by $x(t) = A\sin\omega t+B\cos\omega t$ with $\omega=\sqrt{\frac{K}{m}}$.
Suppose that at time $t = 0$, the position of mass is $x(0)$ and velocity $v(0)$, then its displacement can also be represented as $x(t) = C\cos(\omega t-\phi)$, where $C$ and $\phi$ are
A particle is subjected to two simple harmonic motions as :
$x_1 = \sqrt{7} \sin 5t \text{ cm}$ and $x_2 = 2\sqrt{7} \sin \left(5t + \frac{\pi}{3}\right) \text{ cm}$
where x is displacement and t is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2} \text{ ms}^{-2}$. The value of x is :