Question

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 
 

• A. \( \frac{\sqrt{k r^2}}{\sqrt{J - mr^2}} \)
• B. \( \frac{\sqrt{k r^2}}{\sqrt{J + mr^2}} \)
• C. \( \sqrt{\frac{k}{m}} \)
• D. \( \frac{\sqrt{k r^2}}{J} \)

Hint:

For mechanical systems involving masses, pulleys, and springs, the natural frequency can be found by considering both translational and rotational motion, and solving the system using Newton's second law.

Answer 1

The Correct Option is B

We are given the system with mass \( m \), spring stiffness \( k \), a pulley with radius \( r \), and a moment of inertia \( J \). The mass vibrates harmonically about its static equilibrium position, and we are tasked with finding the natural frequency of the system. Step 1: Write the equation of motion For a system with a spring and mass attached to a rotating pulley, we can derive the equation of motion by considering both the translational motion of the mass and the rotational motion of the pulley. For the spring force, we have: \[ F_{\text{spring}} = -k x, \] where \( x \) is the displacement of the mass. For the rotational motion of the pulley, the torque is related to the angular displacement \( \theta \), and we have: \[ \tau = J \alpha, \] where \( \alpha \) is the angular acceleration of the pulley. The relationship between the angular acceleration \( \alpha \) and the linear acceleration \( a \) of the mass is: \[ a = r \alpha. \] Step 2: Apply Newton's second law Using Newton's second law for the mass \( m \) and for the pulley, we have the following equation for the motion of the system: \[ m \ddot{x} = -k x + mr^2 \ddot{\theta}, \] and the rotational equation for the pulley is: \[ J \ddot{\theta} = -mr^2 \ddot{x}. \] Substituting \( \ddot{x} = r \ddot{\theta} \) into the equation for the mass gives: \[ m r \ddot{\theta} = -k x + mr^2 \ddot{\theta}. \] Now, substitute \( x = r \theta \) to get the system in terms of \( \theta \): \[ m r^2 \ddot{\theta} = -k r \theta + J \ddot{\theta}. \] Step 3: Solve for the natural frequency The equation is now in the form of a standard harmonic oscillator: \[ \ddot{\theta} + \frac{k r}{m r^2 + J} \theta = 0. \] The natural frequency \( \omega \) is given by: \[ \omega = \sqrt{\frac{k r}{m r^2 + J}}. \] Thus, the natural frequency of the system is: \[ \boxed{\omega = \frac{\sqrt{k r^2}}{\sqrt{J + mr^2}}}. \] Hence, the correct answer is Option (B).
Final Answer: (B) \( \frac{\sqrt{k r^2}}{\sqrt{J + mr^2}} \)