Question

Help me determine the restraining force for an elastic pendulum. The problem is the following A particle of massm 'm' is suspended by a massless spring of length 'L'. It hangs, without initial motion, in a gravitational field of strength 'g'. It is struck by an impulsive horizontal blow, which introduces an angular velocity 'ω'. If 'ω' is sufficiently small, it is obvious that the mass moves as a simple pendulum. If 'ω' is sufficiently large, the mass will rotate about the support. Use a Lagrange multiplier to determine the conditions under which the string becomes slack at some point in the motion.

Answer 1

The restraining force for an elastic pendulum is given by the equation F = -(mg/x)x, where F is the force, m is the mass, g is the acceleration due to gravity, and x is the displacement from equilibrium.

Step-by-step explanation:

A simple pendulum consists of a mass m suspended by a string of length L. When the pendulum is displaced from its equilibrium position, a restoring force is exerted by the string, causing the mass to oscillate back and forth. The magnitude of the restoring force can be determined using Hooke's law, which states that the force is directly proportional to the displacement from equilibrium. In this case, the restoring force can be calculated using the equation F = -kx, where k is the spring constant and x is the displacement of the mass.

The spring constant k can be found by applying Newton's second law (F = ma) to the mass in the pendulum. Since the only force acting on the mass is the gravitational force (mg), the equation becomes -kx = mg. Solving for k, we find that k = mg/x.

Therefore, the restraining force for an elastic pendulum is given by the equation F = -(mg/x)x.