Question

a cubical body of relative density , S side L floats in water. If it is slightly displaced downwards and released, the frequency. of resulting SHM will be

Answer 1

If the body is displaced downwards by a distance x then restoring force is developed in it which is balanced by the upthrust of the liquid, i.e.,

$U=-kx$

But the upthrust is the weight of the displaced water. Then,

$-\rho\cdot L^2x\cdot g=-kx\\\\\\\rho L^2g=k$

where $\rho$ is the density of the water and $-L^2x$ is nothing but the volume of the displaced water due to the body of side L, so that base area L^2.

Well, the weight of the body,

$mg=S\rho\cdot L^3g$

where $m$ is the mass of the body. Then,

$mg=SL\cdot\rho L^2g\\\\\\mg=SLk\\\\\\\dfrac {k}{m}=\dfrac {g}{SL}$

Then, frequency,

$\nu=\dfrac {1}{2\pi}\sqrt{\dfrac {k}{m}}\\\\\\\underline {\underline {\nu=\dfrac {1}{2\pi}\sqrt{\dfrac {g}{SL}}}}$

Hence (3) is the answer.

Answer 2

The frequency of resulting SHM is $\dfrac{1}{2\pi}\sqrt{\dfrac{g}{SL}}$

(3) is correct option

Explanation:

Given that,

Relative density = S

Side = L

If it is slightly displaced downwards by a distance x .

Then the restoring force is developed in it which is balaqnced by the upthrust force

$F=-kx$

$-\rho L^3 x g=-kx$

$-(\rho L^3 g)x=-kx$

The spring factor is

$k=\rho L^2 g$

Now, The weight of the body

$mg=S\rho L^3 g$

$mg=SL\times\rho L^2 g$

$mg=SLk$

$\dfrac{k}{m}=\dfrac{g}{SL}$

We need to calculate the frequency

Using formula of frequency

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Put the value into the formula

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{SL}}$

Hence, The frequency of resulting SHM is $\dfrac{1}{2\pi}\sqrt{\dfrac{g}{SL}}$

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Topic : frequency

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