(A) When the block is in equilibrium, the weight of the block is balanced by the buoyant force: $mg = V_{submerged} \rho g = (A \cdot h) \rho g$, wher

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$\frac{1}{2\pi} \sqrt{\frac{A\rho g}{m}}$

(A) When the block is in equilibrium, the weight of the block is balanced by the buoyant force: $mg = V_{submerged} \rho g = (A \cdot h) \rho g$, where '$h$' is the submerged depth.When the block is displaced vertically by a small distance '$x$',the additional buoyant force acting on it is $F_{buoyant_extra} = (A \cdot x) \rho g$.This force acts as a restoring force: $F = - (A \rho g) x$.Comparing this with the simple harmonic motion equation $F = -kx$, we get the spring constant $k = A \rho g$.The angular frequency of oscillation is $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{A \rho g}{m}}$.The frequency of oscillation is $f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{A \rho g}{m}}$.

Class 11 Physics Oscillations Method to determine Time Period and Frequency for diffrent type of Object of SHM

Difficult

$A$ rectangular block of mass '$m$' and cross-sectional area '$A$' floats on a liquid of density '$\rho$'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ($g$ = acceleration due to gravity).

MHT CET 2026 All MHT CET

A
$\frac{1}{2\pi} \sqrt{\frac{A\rho g}{m}}$
B
$\frac{1}{2\pi} \sqrt{\frac{m}{A\rho g}}$
C
$2\pi \sqrt{\frac{m}{A\rho g}}$
D
$2\pi \sqrt{\frac{A\rho g}{m}}$

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Method to determine Time Period and Frequency for diffrent type of Object of SHM

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$A$ large horizontal surface moves up and down in $S.H.M.$ with an amplitude of $1 \, cm$. If a mass of $10 \, kg$ placed on the surface is to remain continuously in contact with it,the maximum frequency of $S.H.M.$ will be .... $Hz$.

MediumAIIMS 2007

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An air chamber of volume $V$ has a neck of cross-sectional area $a$ into which a ball of mass $m$ just fits and can move up and down without any friction (Figure). Show that when the ball is pressed down a little and released,it executes $SHM$. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.

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In both figures shown below, there is a hole along the diameter of the Earth. In the first figure, a particle is released from point $A$ and it oscillates with a time period $T_1$. In the second figure, the same particle is released from point $B$ and it oscillates with a time period $T_2$. Then, [$O$ is the centre of the Earth]

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$A$ body of mass $0.01 \ kg$ executes simple harmonic motion $(S.H.M.)$ about $x = 0$ under the influence of a force $F$ as shown in the graph (where $F$ is in $N$ and $x$ is in $m$). The period of the $S.H.M.$ is ... $s$. (Given: The graph is a straight line passing through the origin with slope $k = 4 \ N/m$ calculated from $F = 8 \ N$ at $x = 2 \ m$).

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$A$ particle of mass $m$ undergoes oscillations about $x=0$ in a potential given by $V(x) = \frac{1}{2} k x^2 - V_0 \cos \left(\frac{x}{a}\right)$,where $V_0, k, a$ are constants. If the amplitude of oscillation is much smaller than $a$,the time period is given by

KVPY 2010

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$A$ rectangular block of mass '$m$' and cross-sectional area '$A$' floats on a liquid of density '$\rho$'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ($g$ = acceleration due to gravity).

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$A$ large horizontal surface moves up and down in $S.H.M.$ with an amplitude of $1 \, cm$. If a mass of $10 \, kg$ placed on the surface is to remain continuously in contact with it,the maximum frequency of $S.H.M.$ will be .... $Hz$.

$A$ particle of mass $m$ undergoes oscillations about $x=0$ in a potential given by $V(x) = \frac{1}{2} k x^2 - V_0 \cos \left(\frac{x}{a}\right)$,where $V_0, k, a$ are constants. If the amplitude of oscillation is much smaller than $a$,the time period is given by

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