13.1 Describing oscillationsThe spring drives the glider back and forth on the air-track and you can observe the changes in the free-body diagram as the motion proceeds from A to A and back.

equilibrium position:Fnet = 0a = 0v is maxKE is maxPEs is minRestoring force

E = PEs + K = constantCharacteristics of periodic motionFs = - kx

The amplitude of the motion, denoted by A, is the maximum magnitude of displacement from equilibrium or the maximum of |x|.The period, T, is the time for one cycle. It is always positive. Its SI unit is second.The frequency, f, is the number of cycles in a unit of time. It is always positive, its SI units is hertz: 1 Hz= 1 s-1The angular frequency, , is 2 times the frequency: = 2fThe relationships between frequency and period:f = 1/TT = 1/fFor angular frequency: = 2f = 2/TAmplitude, Period, Frequency, and Angular Frequency

Example 13.1 Period, frequency, and angular frequencyAn ultrasonic transducer used for medical diagnosis oscillates at a frequency of 6.7 MHz. How much time does each oscillation take, and what is the angular frequency?

For each of the following values of the bodys x-velocity vx and x-acceleration ax, state whether its displacement x is positive, negative, or zero.vx > 0 and ax > 0vx > 0 and ax < 0vx < 0 and ax > 0vx < 0 and ax < 0vx = 0 and ax < 0vx > 0 and ax = 0 negativepositivex = 0x < 0x > 0zeronegativepositivepositiveTest Your Understanding 13.1

When the restoring force Fx is directly proportional to the displacement from equilibrium x, such as:

The minus sign means the acceleration and displacement always have opposite signs.This acceleration is not constant, so we cant use the constant acceleration equations.13.2 Simple harmonic motionthe oscillation is called simple harmonic motion, or SHM.

Object moving on an ideal spring, the motion is SHM

Uniform circular motionSimple harmonic motion viewed as a projection of uniform circular motionSHM

Displacement of SHMx = AcosSince Rotation is Uniform, is constant. = tx = Acost

velocityx = Acostv = A(-sint)()v = -Asint

Acceleration v = -Asinta = -A(cost)()a = -A2costa = - 2x

The acceleration of the point P is directly proportional to the displacement x and always has the opposite sign hallmarks of simple harmonic motion.a = - 2x

Comparing equations:

We get: Note: is the angular speed of Q (circular motion) and the angular frequency of SHM. These quantities are equal!When you start a body oscillating in SHM, the value of angular frequency is not yours to choose; it is predetermined by the values of k and m. the units k are N/m or kg/s2, so the unit of is rad/s.

Correspondingly, we have frequency and period as:The equation shows that an object with a larger mass m, greater inertia, will have less acceleration, move less frequently, take a longer time for a complete cycle. In contrast, a stiffer spring (one with greater constant k) exerts a greater force at a given deformation x, causing greater acceleration, higher frequency, and a shorter time T per cycle.The period T and frequency f of SHM are completely determine by m and k, it does not depend on the amplitude A.Large A means larger restoring forces, larger average speed to compensate for having to travel a larger distance, so the same total time is involved.

Example 13.2 angular frequency, frequency, and period in SHMA spring is mounted horizontally, with its left end held stationary. By attaching a spring balance to the free end and pulling toward the right, we determine that the stretching force is proportional to the displacement and that a force 6.0 N causes a displacement of 0.300 m. We remove the spring balance and attach a 0.50 kg glider to the end, pull it a distance of 0.020 m along a frictionless air track, release it, and watch it oscillate.

Find the force constant of the spring.Find the angular frequency, frequency, and period of the oscillation

x versus t graph

SHM phase, position, velocity, and accelerationSHM can occur with various phase angles.For a given phase we can examine position, velocity, and acceleration.

Watch variables change for a glider example As the glider undergoes SHM, you can track changes in velocity and acceleration as the position changes between the classical turning points.

A glider is attached to a spring. If the glider is moved to x = 0.10 m and released from rest at time t = 0, it will oscillate with amplitude A = 0.10 m and phase angle = 0.Suppose instead that at t = 0 the glider is at x = 0.10 m and is moving to the right. In this situation Is the amplitude greater than, less than, or equal to 0.10 m?Is the phase angle greater than, less than, or equal to zero?Suppose instead that at t = 0 the glider is at x = 0.10 and is moving to the left. In this situationIs the amplitude greater than, less than, or equal to 0.10 m?Is the phase angle greater than, less than, or equal to zero?

Test Your Understanding 13.2

The force exerted by an ideal spring is a conservative force, and the vertical forces do no work, so the total mechanical energy of the system of a body in SHM is conserved. (assume that the mass of the spring itself is negligible.

When x = A (or A), vx = 0. at this point the energy is entirely potential, E = kA2. Because E is constant, it is equal to kA2 at any point13.3 Energy in SHMUsing this equation, we can solve for vx at any given point x if A is known.

Interpreting E, K, and U in SHMEnergy is conserved during SHM and the forms (potential and kinetic) interconvert as the position of the object in motion changes.

Example 13.4 Velocity, acceleration, and energy in SHMIn the oscillation described in Example 13.2, k = 200 N/m, m = 0.50 kg, and the oscillating mass released from rest at x = 0.020 m.Find the maximum and minimum velocities attained by the oscillating body.Compute the maximum acceleration.Determine the velocity and acceleration when the body has moved halfway to the center from its original position.Find the total energy, potential energy, and kinetic energy at this position.

Example 13.5 Energy and momentum in SHMA block with mass M attached to a horizontal spring with force constant k is moving with simple harmonic motion having amplitude A1. At the instant when the block passes through its equilibrium position, a lump of putty with mass m is dropped vertically onto the block from a very small height and sticks to it.Find the new amplitude and period.Repeat part (a) for the case in which the putty is dropped on the block when it is at one end of its path.

To double the total energy for a mass-spring system oscillating in SHM, by what factor must the amplitude increase?4221/2 21/4 By what factor will the frequency change due to this amplitude increase?4221/2 21/4 It does not change

Test Your Understanding 13.3

SHM can occur in any system in which there is a restoring force that is directly proportional to the displacement from equilibrium, as given by: Fx = - kx13.4 Applications of Simple Harmonic MotionVertical SHM doesnt differ in any essential way from horizontal SHM. The only real change is the equilibrium position.The same ideas hold if a body with weight mg is placed atop a compressible spring.Vertical SHM

Example 13.6 Vertical SHM in an old carThe shock absorbers in an old car with mass 1000 kg are completely worn-out. When a 980 N person climbs slowly into the car to its center of gravity, the car sinks 2.8 cm. when the car, with the person aboard, hits a bump, the car starts oscillating up and down in SHM. Model the car and person as a single body on a single spring, and find the period and frequency of the oscillation.

Watches keep time based on regular oscillations of a balance wheel initially set in motion by a spring.The wheel has a moment of inertial I about its axis. A coil spring exerts a restoring torque z that is proportional to the angular displacement from the equilibrium position.z = - (analog to Fx = -kx)Where (kappa) is a constant called torsion constant

Angular SHMThe motion of rotational SHM is described by the function: = cos(t + )

A block attached to a hanging ideal spring oscillates up and down with a period of 10 s on earth. If you take the block and spring to Mars. Where the acceleration due to gravity is only about 40% as large as on earth, what will be the new period of oscillation?10 sMore than 10 sLess than 10 sTest Your Understanding 13.4

A simple pendulum is an idealized model consisting of a point mass suspended by a massless, unstretchable string. When the point mass is pulled to one side of its straight-down equilibrium position and released, it oscillates about the equilibrium position.A person a swing can be modeled as simple pendulum13.5 The Simple Pendulum

The restoring force is provided by gravity; The motion is approximately simple harmonic.If the is small, sin

graphs

Length (m)periodPeriod vs. lengthperiodPeriod vs.

Example 13.8 A simple pendulumFind the period and frequency of a simple pendulum 1.000 m long at a location where g = 9.800 m/s2.

exampleA simple pendulum of length l, whose bob has mass m, oscillates with a period T. If the bob is replaced by one of mass 4m, what is the period of oscillation in terms of T?

exampleA pendulum with a period of 1 s on Earth, where the acceleration due to gravity is g, is taken to another planet, where its period is 2 s. What is the acceleration due to gravity on the other planet?

exampleA simple pendulum has a period of 2 s for small amplitude oscillations. What is the length of the pendulum?

When a body oscillating on a horizontal spring passes through its equilibrium position, its acceleration is zero. When the bob of an oscillating simple pendulum pass through its equilibrium position, is its acceleration zero?No, because there is a net force (centripetal force) acting on the bob causing a change in direction of motion.Test Your Understanding 13.5

Damped oscillationsA person may not wish for the object they study to remain in SHM. Consider shock absorbers and your automobile. Without damping the oscillation, hitting a pothole would set your car into SHM on the springs that support it.

When the body is displaced as shown, the weight mg causes a restoring torque:A physical pendulum is any real pendulum that uses an extended body in motion. When is small, we can approximate that:13.6 The Physical Pendulum

Example 13.9 Physical pendulum vs. simple pendulumSuppose the body in the figure is a uniform rod with length L, pivoted at one end. Find the period of its motion.

The center of gravity of simple pendulum of mass m and length L is located at the position of the pendulum bob, a distance L from the pivot point. The center of gravity of a uniform rod of the same mass m and length 2L pivoted at one end is also a distance L from the pivot point. How does the period of this uniform rod compare to the period of the simple pendulum?The rod has a longer periodThe rod has a shorter periodThe rod has the same periodI = 1/3 m(2L)2 = 4/3 mL2d = LTest Your Understanding 13.6

Lab determine spring constant KPurpose: Use SHM to determine spring constant kMaterial: (questions to ask your self: What data do you need to determine k? How are you going to obtain your data? How are you going to minimize experimental error?)Procedure: Briefly describe how the lab is going to be done. Someone who was not present during the lab should be able to understand how the experiment was perforem and be able to reproduce the results by reading your procedure.Data table: record you data with units in the headingsData analysis: use your collected data to determine spring constant.

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