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Vibrations and Springs The force of a spring on a mass varies with the displacement, x, as referenced from the initial position, x0. (a) When x is positive (mass pulled to right for this figure), the spring is in tension and the restoring force, Fs, is to the left. (b) Spring in equilibrium at x0. Restoring force, Fs, is zero. (c) When x is negative (mass pushed to left for this figure), the spring is in compression and the restoring force, Fs, is to the right. Vibrations, springs, and Hooke’s Law The restoring force for springs was discovered by Robert Hooke in 1678 and is know as Hooke’s Law. Fs = - kx k is a positive constant know as the spring stiffness x is the displacement as measure from the equilibrium position (sign convention: Fs is in the opposite direction of x, therefore “-” sign) Harmonic Motion Simple harmonic motion is the study of oscillations. An oscillation is motion of an object that regularly repeats itself over the same path. Examples: pendulum, mass vibrating on spring Here, the restoring force is generated by gravity trying to put the pendulum back in equilibrium (vertical position). Here, Fs = -kx Harmonic Motion An object is said to be in simple harmonic motion if the following occurs: •It moves in a straight line. •A variable force acts on it. •The magnitude of force is proportional to the displacement of the mass. •The force is always opposite in direction to the displacement direction. •The motion is repetitive and a round trip, back and forth, is always made in equal time periods. Definition of terms for simple harmonic motion: restoring force, Fs– force that restores an object to the equilibrium position Cycle– one full repetition of periodic motion Amplitude, A – maximum displacement from the equilibrium position during a cycle Period, T – time it takes for an object to pass through one complete cycle Frequency, f– number of cycles per unit of time Period and frequency have an inverse relationship: The spring constant, k For a vertical spring-mass system, the equilibrium point is when the restoring force, Fs, pulling upwards is in equilibrium with the weight of the mass pulling down, W = mg. Using Hooke’s Law: Fs = kx = mg k = mg/x In this case the acceleration of the mass is due to gravity. Newton’s Law and Hooke’s Law The springs we will study here exhibit a linear restoring force, Fs. This spring force is a conservative force, and is a form of potential energy. As a object moves in simple harmonic motion, we can use Newton’s 2nd Law (F = ma) to calculate the acceleration of the mass in motion. Equating Newton’s and Hooke’s Laws: -kx = ma Since amplitude, A, is maximum or minimum displacement, the range in acceleration over the full harmonic cycle is: -kA/m= +kA/m Elastic Potential Energy Elastic potential energy is Potential energy stored as a result of deformation of an elastic object, such as the stretching of a spring. It is equal to the work done to stretch the spring. Force has the form: Fs = -kx Work done to stretch the spring a distance x: Elastic potential energy is proportional to the spring constant, k, and the square of the displacement, x2. The system has zero PE at the equilibrium position. Harmonic Motion - Conservation of Energy In a mass in harmonic motion, three types of energy act on the system: 1) Potential energy from the spring as it tries to restore the system to equilibrium 2) Gravitational potential energy 3) Kinetic energy of mass as it moves At the maximum displacement (amplitude, A), the initial energy stored in the system: PEinitial = kA2/2 After the spring is released, the energy of the system in harmonic motion is both the elastic PE plus the KE: PE + KE = kx2/2 + mv2/2 For equilibrium: kA2/2 = kx2/2 + mv2/2 Relationship of velocity to position: v = ±{k(A2 –x2)/m}1/2 Example – 14.4 A 3 kg mass is attached to a spring and pulled out horizontally to a maximum displacement from equilibrium of 0.5m. What spring constant must the spring have if the mass is to achieve an acceleration equal to that of gravity? Example – 14.19 A mass-spring system oscillates with an amplitude of 3.5 cm. If the spring constant is 250 N/m and the mass is 0.5 kg, determine: (a) The mechanical energy of the system Example – 14.19 A mass-spring system oscillates with an amplitude of 3.5 cm. If the spring constant is 250 N/m and the mass is 0.5 kg, determine: (b) The maximum speed of the mass Example – 14.19 A mass-spring system oscillates with an amplitude of 3.5 cm. If the spring constant is 250 N/m and the mass is 0.5 kg, determine: (c) The maximum acceleration of the mass Example – 14.61 A mass of 1.5 kg is attached to a spring of spring constant 1000 n/m. Find the force on the mass and its acceleration at the positions (equilibrium is at x = 0): (a) 0.5 m (b) 0.1 m (c) 0 m (d) -0.2m (e) 0.5m