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Question: A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its lengt...

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 r 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 oscillation of the cylinder is –

A

12π(kAρgM)1/2\frac{1}{2\pi}\left( \frac{k–A\rho g}{M} \right)^{1/2}

B

12π(k+AρgM)1/2\frac{1}{2\pi}\left( \frac{k + A\rho g}{M} \right)^{1/2}

C

12π(k+ρgL2M)1/2\frac{1}{2\pi}\left( \frac{k + \rho gL^{2}}{M} \right)^{1/2}

D

12π(k+AρgAρg)1/2\frac{1}{2\pi}\left( \frac{k + A\rho g}{A\rho g} \right)^{1/2}

Answer

12π(k+AρgM)1/2\frac{1}{2\pi}\left( \frac{k + A\rho g}{M} \right)^{1/2}

Explanation

Solution

F = – {kY + AYrg} Ž Ma = – {k + Arg} Y

a = – {k+AρgM}Y\left\{ \frac{k + A\rho g}{M} \right\} Y Ž f = 12π\frac { 1 } { 2 \pi } (K+Aρg)/M\sqrt{(K + A\rho g)/M}