For the system given below,find the angular f | vedclass

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(C) The system consists of a block of mass $m = 1/4 	ext{ kg}$ attached to a string passing over a disc of mass $M = 1 	ext{ kg}$ and radius $R$,wit

Vedclass · Accepted Answer

$\frac{20}{\sqrt{3}}$

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(C) The system consists of a block of mass $m = 1/4 	ext{ kg}$ attached to a string passing over a disc of mass $M = 1 	ext{ kg}$ and radius $R$,with the other end of the string attached to a spring of constant $K = 100 	ext{ N/m}$.Let $x$ be the displacement of the block. The velocity of the block is $v = \dot{x}$. The angular velocity of the disc is $\omega_d = v/R$. The extension in the spring is $x$,so the spring force is $Kx$.The total energy $E$ of the system is given by:$E = \frac{1}{2} Kx^2 + \frac{1}{2} mv^2 + \frac{1}{2} I \omega_d^2 = 	ext{constant}$For a disc,the moment of inertia is $I = \frac{1}{2} MR^2$. Substituting this and $\omega_d = v/R$:$E = \frac{1}{2} Kx^2 + \frac{1}{2} mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Kx^2 + \frac{1}{2} mv^2 + \frac{1}{4} Mv^2$$E = \frac{1}{2} Kx^2 + \frac{1}{2} (m + \frac{M}{2}) v^2$Differentiating with respect to time $t$:$\frac{dE}{dt} = Kx \dot{x} + (m + \frac{M}{2}) v \dot{v} = 0$Since $\dot{x} = v$ and $\dot{v} = a$:$Kxv + (m + \frac{M}{2}) va = 0$$a = -\frac{K}{m + M/2} x$Comparing with $a = -\omega^2 x$,the angular frequency is:$\omega = \sqrt{\frac{K}{m + M/2}}$Substituting the given values $K = 100 	ext{ N/m}$,$m = 0.25 	ext{ kg}$,$M = 1 	ext{ kg}$:$\omega = \sqrt{\frac{100}{0.25 + 1/2}} = \sqrt{\frac{100}{0.75}} = \sqrt{\frac{100}{3/4}} = \sqrt{\frac{400}{3}} = \frac{20}{\sqrt{3}} 	ext{ rad/s}$.

For the system given below,find the angular frequency of oscillation.

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