Consider the system shown in the figure. A rope goes over a pulley. A mass, m is hanging from the rope. A spring of stiffness, k is attached at one end of the rope. Assume rope is inextensible, massless and there is no slip between pulley and rope.

The pulley radius is r and its mass moment of inertia is J. Assume that the mass is vibrating harmonically about its static equilibrium position. The natural Frequency of the system is

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GATE ME 2021 Official Paper: Shift 2

Option 1 : \(\sqrt{\frac{kr^2}{J + mr^2}}\)

CT 3: Building Materials

3014

10 Questions
20 Marks
12 Mins

**Explanation:**

**Apply energy method:**

Let system is displaced by a small angle 'θ'

Since the system is in static equilibrium position

So net energy E, and **\( \frac{dE}{dt} =0\)**

Energy of spring = \(\Rightarrow U = \frac{1}{2}kx^2 \)

Energy of pulley during rotation \(\Rightarrow U = \frac{1}{2}Jω^2 \)

Energy of bucket mass m \(\Rightarrow U = \frac{1}{2}mv^2 \)

Net energy

\(E= \frac{1}{2}kx^2 +\frac{1}{2}Jω^2 +\frac{1}{2}mv^2\)

x = rθ , w = \(\dot{\theta}\), v = rω

\(E= \frac{1}{2}kr^2θ^2 +\frac{1}{2}J \dot{θ} ^2 +\frac{1}{2}mr^2w^2\)

\(E= \frac{1}{2}kr^2θ^2 +\frac{1}{2}J \dot{θ} ^2 +\frac{1}{2}mr^2\dot{θ}^2\)

\(\frac{dE}{dt}= \frac{1}{2}kr^2(2θ)\dot{\theta} +\frac{1}{2}J (2\dot{θ})\ddot\theta +\frac{1}{2}mr^2(2\dot{θ})\ddot\theta = 0\)

\(\frac{dE}{dt}= kr^2θ \dot{\theta} +J \dot{\theta}\ddot\theta +mr^2\dot{\theta}\ddot\theta =0\)

\(\frac{dE}{dt}= kr^2θ +J \ddot\theta +mr^2\ddot\theta =0\)

\(\frac{dE}{dt}= kr^2θ +(J +mr^2)\ddot\theta =0\)

\( kr^2θ +(J +mr^2)\ddot\theta =0\)

\(\ddot\theta+ \frac{kr^2θ} {(J +mr^2)} =0\)

Comparing \(\ddot\theta+ {\omega^2_n\theta} {} =0\)

\(\omega^2_n = \frac{kr^2} {(J +mr^2)} \)

\(\omega_n=\sqrt{\frac{kr^2}{J+ mr^2}}\)