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Question

A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density ρ at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is 'k', the frequency of oscillations of the cylinder is?

A
12π(kAρgM)1/2
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B
12π (k+AρgM)1/2
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C
12π(k+ρgLM)1/2
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D
12π(k+AρgAρg)1/2
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Solution

The correct option is C 12π (k+AρgM)1/2

When the cylinder is pushed and released then, the force applied and the up thrust force will be equal and it can be written as,

F=(kx+Axρg)

The acceleration is given as,

a=(kx+Axρg)M

The frequency of the oscillation is given as,

f=12πax

=12π(k+Aρg)MHz

Thus, the frequency of the oscillation of the cylinder is 12π(k+Aρg)MHz.


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