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### Syllabus

### Goals for Chapter 13

### 13.3Energy in Simple Harmonic Motion

### 13.3Energy in Simple Harmonic Motion

13.1Describing Oscillation

13.2Simple Harmonic Motion

13.3Energy in Simple Harmonic Motion

13.4Applications of Simple Harmonic Motion

13.5The Simple Pendulum

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

To follow periodic motion to study simple harmonic motion.

To use the pendulum as a prototypical system undergoing simple harmonic motion.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Many kinds of motion repeat themselves over and over: the vibration of a quartz crystal in a watch, the swinging pendulum of a grandfather clock, and the back-and-forth motion of the pistons in a car engine.### Introduction

- We call this periodic Motion or Oscillation, and it’s the subject of this chapter.

- Understanding periodic motion will be essential for later study of waves, sound, and light.

- Body undergoes periodic motion always has a stable equilibrium position. When moves away from this position and released, Force pull it back toward equilibrium.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

One of the simplest system that can have periodic motion is shown in figure bellow.### 13.1Describing Oscillation

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The spring force is the only horizontal force acting on the body; the vertical normal and gravitational forces always add to zero.### 13.1Describing Oscillation

- The quantities x, vx, ax, and Fx refer to the x-components of the position, velocity, acceleration, and the force vectors, respectively.

- In our coordinate system, the origin O is at the equilibrium position, where the spring neither stretched or compressed.

- Then x is the x-component of the displacement of the body from equilibrium and is also the change in the length of the spring.

- Then x-component of acceleration is given by ax = Fx/m.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Whenever the body is displaced from its equilibrium position, the spring force tends to restore it to the equilibrium position.### 13.1Describing Oscillation

- We call a force with this character a restoring force.

- Oscillation can occur only when there is a restoring force tending to return the system to equilibrium.

- If we displace the body to the right to x = A and then let it go, the net force and the acceleration to the left.

- The speed increases as the body approaches the equilibrium position O.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

When the body is at O, the net force acting it is zero, but because of its motion it overshoots the equilibrium position.### 13.1Describing Oscillation

- On the other side of the equilibrium position the velocity to the left but the acceleration is to the right; the speed decreases until the body comes to a stop.

- The body is oscillating! If there is no friction or other force to remove mechanical energy from the system, this motion repeat itself forever.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Here are some terms that we will use in discussing periodic motion of all kinds:### 13.1Describing Oscillation

- The amplitude of the motion, denoted by A, is the maximum magnitude of displacement from equilibrium – that is, the maximum value of |x|. It is always positive. The SI unit of A is meter.

- The period, T, is the time of one cycle. It is always positive. The SI unit is second.

- The frequency, f, is the number of cycles in a unit of time. It is always positive. The SI unit is hertz.

1 hertz = 1 Hz = 1 cycle/s = 1 s-1

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

(relationships between frequency and period)### 13.1Describing Oscillation

- The angular frequency, ω, is 2π times the frequency, its unit is rad/s:

- From the definitions of period T and frequency f,

- Also from the definition of ω,

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Example 13.1( Period, frequency, and angular frequency):### 13.1Describing Oscillation

An ultrasonic transducer (a kind of loudspeaker) used for medical diagnosis oscillators at a frequency of 6.7 MHz = 6.7 × 106 Hz. How much time does oscillation take, and what is the angular frequency?

Solution:

- Identify and Set Up:Our target variables are the period T and the angular frequency ω. We can use Eqs. (13.1) & (13.2).

- Execute: From Eqs. (13.1) & (13.2),

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The simplest kind of oscillation occur when the restoring force Fx is directlyproportional to the displacement from equilibrium x.### 13.2Simple Harmonic Motion

- This happens if the spring is an ideal one that obeys Hooke’s law.

- The constant of proportionality between Fx and x is the force constantk.

- On either side of the equilibrium position, Fx and x have opposite signs. The x-component of force Fx on the body is:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Last equation gives the correct magnitude and sign of the force, whether x is positive, negative, or zero.### 13.2Simple Harmonic Motion

- The force constant is always positive and has unit of N/m.

- When the restoring force is directly proportional to the displacement from equilibrium, as given by Eq. (13.3), the oscillation is calledsimple harmonic motion, SHM.

- The acceleration ax = d2x/dt2 = Fx/m of a body in SHM is given by

This acceleration is not constant

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

A body that undergoes simple harmonic motion is called a harmonic oscillator.### 13.2Simple Harmonic Motion

- Keep in mind that not all periodic motion are simple harmonic.

Equation of Simple Harmonic Motion

- The angular frequency of SHM,

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The frequency f and period T are,### 13.2Simple Harmonic Motion

- From Eq. (13.12) a large mass m, with its greater inertia, will have less acceleration, move more slowly, and take a longer time to a complete cycle.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Example 13.2(Angular frequency, frequency, and period in SHM):### 13.2Simple Harmonic Motion

A spring is mounted horizontally, with its left end held stationary. By attaching a spring balance to the free end and pulling toward the right, figure below, we determine that the stretching force is proportional to the displacement and that a force 6.0 N causes a displacement of 0.030 m. We remove the spring balance and attach a 0.50-kg body to the end, pull it a distance of 0.020 m, release it, and watch it oscillate. a) Find the force constant of the spring. b) Find the angular frequency, frequency, and period of the oscillation.

Solution:

- Identify:Because the spring force is proportional to the displacement, the motion is SH.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Set Up:We use the value of force constant k using Hooke’s law, Eq. (13.3), and ω, f, and T using Eqs. (13.10), (13.11), and (13.12).### 13.2Simple Harmonic Motion

- Execute:a) When x = 0.030 m, the force the spring exerts on the spring balance is Fx = - 6.0 N. From Eq. (13.3),

b) Using m = 0.50 kg in Eq. (13.10), we find:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The frequency f is:### 13.2Simple Harmonic Motion

- The period T is:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Displacement, Velocity, and acceleration in SHM### 13.2Simple Harmonic Motion

- We will find the displacement x as a function of time for a harmonic oscillation.

- The constant Ø in Eq. (13.13) is called phase angle.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

We denote the position at t = 0 by xo. Putting t = 0 and x = xo in Eq. (13.13), we get### 13.2Simple Harmonic Motion

- To find the velocity vx and acceleration ax as a function of t for a harmonic oscillator taking derivatives of Eq. (13.13) with respect to t

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The velocity vx oscillate between vmax= + ωA and – vmax= - ωA.### 13.2Simple Harmonic Motion

- The acceleration ax oscillate between amax= +ω2A and - amax= -ω2A.

- Comparing Eq. (13.16) with Eq. (13.13) and recalling that ω2 = k/m from Eq. (13.10), we see that

- When the body is passing through the equilibrium position so that the displacement is zero, the velocity equals either vmax or –vmax (depending on which way the body is moving) and the acceleration is zero.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

When the body is at either its maximum positive displacement, x = +A, or its maximum negative displacement, x = -A, the velocity is zero and the body is at instantaneously at rest.### 13.2Simple Harmonic Motion

- At these points, the restoring forceFx= - kx and the acceleration of the body have their maximum magnitudes.

- At x = +A the acceleration is negative and equal to – amax. At x = -A the acceleration is positive ax= + amax.

- If we are given the initial position xo and initial velocity vox for the oscillating body, we can determine the amplitude A and the phase angle Ø.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The initial velocity vox is the velocity at time t = 0; putting vx = vox and t = 0 in Eq. (13.15), we find: ### 13.2Simple Harmonic Motion

- To find Ø, divide Eq. (13.17) by Eq. (13.14). This eliminates A and gives an equation that we can solve for Ø:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The amplitude A is:### 13.2Simple Harmonic Motion

- Graphs from a particle undergoing simple harmonic motion.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Example 13.3(Describing SHM):### 13.2Simple Harmonic Motion

Let’s return to the system of mass and horizontal spring we considered in Example 13.2, with k = 200 N/m and m = 0.50 kg. This time we give the body an initial displacement of +0.015 m and an initial velocity of +0.40 m/s a) Find the period, amplitude, and phase angle of the motion. b) Write equations for the displacement, velocity, and acceleration as function of time.

Solution:

- Identify:The oscillations are SHM.

- Set Up:We are given the values of k, m, xo, vox, from them, we calculate T, A, and Ø and the expression for x, vx, and ax as function of time.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Execute:a) The period is exactly the same as in Example 13.2, T = 0.31 s. In SHM the period does not depend on the amplitude, only on the values of k and m.### 13.2Simple Harmonic Motion

- In Example 13.2 we found that ω = 20 rad/s. So from Eq. (13.19).

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

To find the phase angle Ø, we use Eq. (13.18).### 13.2Simple Harmonic Motion

b)The displacement, velocity, and acceleration at any time are given by Eqs (13.13), (13.15), and (13.16), respectively. Substituting the values, we get:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Take another look at the body oscillating on the end of a spring in Fig. 13.1.### 13.3Energy in Simple Harmonic Motion

- The spring force is the only horizontal force on the body.

- 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 is conserved. We assume that the mass of the spring itself is negligible.

- The kinetic energy of the body is K = ½ m v2 and the potential energy of the spring is U = ½ k x2.

- There are no nonconservative forces that do work, so the total mechanical energyE = K + U is conserved:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Since the motion is one-dimensional, v2 = vx2. ### 13.3Energy in Simple Harmonic Motion

- The total mechanical energy E is also directly related to the amplitude A.

- When x = A (or –A), vx= 0. At this point the energy is entirely potential, and E = ½ kA2.

- Because E is constant, this quantity equals E at any other point. Combining this expression with Eq. (13.20), we get

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

- We can use Eq. (13.21) to solve for velocity vx of the body at a given displacement x:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

- The ± sign means that at a given value of x the body can be moving in either direction.

- When x= ± A/2,

- Eq. (13.22) also shows that the maximum speed vmax occurs at x = 0. Using Eq. (13.10), we find that:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Total energy### 13.3Energy in Simple Harmonic Motion

Kinetic energy

Potential energy

O

-A

+A

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Example 13.4(Velocity, acceleration, and energy in SHM):### 13.3Energy in Simple Harmonic Motion

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

Solution:

- Identify:The problem refers to the motion at various positions in the motion, not at specified times. We can use the energy relations.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Set Up:For any position x we use Eqs. (13.22) and (13.14) to find the velocity vx and acceleration ax. Given the velocity and position, we use Eq. (13.21) to find the energy quantities K, U, and E.### 13.3Energy in Simple Harmonic Motion

- Execute:a) The velocity vxat any displacement x is given by Eq. (13.22):

- The maximum velocity occurs when the body is moving to the right through the equilibrium position, where x = 0:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The minimum (i.e., most negative) velocity occurs when the body is moving to the left through x = 0; its value is – vmax = - 0.40 m/s.### 13.3Energy in Simple Harmonic Motion

b) From Eq. (13.4),

- The maximum (most positive) acceleration occurs at the most negative value of x, x = - A; therefore,

- The minimum (most negative) acceleration is - 8.0 m/s2, occurring at x = + A = + 0.020 m.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

c)At a point halfway to the centre form the initial position, x = A/2 = 0.010 m. From Eq. (13.22),### 13.3Energy in Simple Harmonic Motion

- We choose the negative square root because the body is moving from x = A toward x = 0. From Eq. (13.4),

- At this point the velocity and the acceleration have the same sign, so the speed is increasing.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

d)The total energy has the same value at all points during the motion:### 13.3Energy in Simple Harmonic Motion

- The potential energy is:

- and the kinetic energy is:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

We have looked at one situation in which simple harmonic motion occurs: a body attached to an ideal horizontal spring.### 13.4Applications of SHM

- But SHM can occur in any system in which there is a restoring force that is directly proportional to the displacement from equilibrium.

Vertical SHM

- Suppose we hang a spring with force constant k (figure below) and suspend from it a body with mass m. Oscillations will now be vertical; will they still be SHM?

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

l### 13.4Applications of SHM

l

l

Δl - x

F = k (Δl – x)

(a)

Δl

F = k Δl

x

Object displaced from equilibrium: net force is proportional to displacement, oscillations are SHM

x = 0

Object in equilibrium:

(spring force) = (weight force)

mg

(c)

mg

(b)

University Physics, Chapter 13

Vertical SHM

- In the figure below (b) the body hangs at rest, in equilibrium

Dr. Y. Abou-Ali, IUST

Vertical SHM### 13.4Applications of SHM

- In this position the spring is stretched an amount Δl just great enough that the spring’s upward vertical force k Δl on the body balances its weight mg:

- Take x = 0 to be this equilibrium position and take the positive x- direction to be upward.

- When the body is a distance xabove its equilibrium position (figure c), the extension of the spring is Δl – x.

- The upward force it exerts on the body is then k (Δl – x), and the net x-component of force on the body is:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Vertical SHM### 13.4Applications of SHM

- That is, a net downward force of magnitude kx.

- Similarly, when the body is below the equilibrium position, there is a net upward with magnitude kx.

- In either case there is a restoring force with magnitude kx.

- If the body is set in vertical motion, it oscillates in SHM, with angular frequency ω = (k/m) 1/2.

- Vertical SHM does not differ in any way from horizontal SHM.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Vertical SHM### 13.4Applications of SHM

- The only real change is that the equilibrium position x = 0 no longer corresponds to the point at which the spring is unstretched.

- The same ideas hold if a body with weight mg is placed atop a compressed spring and a compresses it a a distance Δl.

Example 13.6(Vertical SHM in an old car):

The 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 centre 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 the frequency of the oscillation.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

F### 13.4Applications of SHM

Δl

mg

Vertical SHM

Solution:

- Identify and Set Up: The situation is like that shown in figure below.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Vertical SHM### 13.4Applications of SHM

- Execute:When the force increases by 980 N, the spring compresses an additional 0.028 m, and the coordinate x of the car changes by – 0.028 m.

- Hence the effective force constant (including the effect of the entire suspension) is:

- The person’s mass is w/g = (980 N)/(9.8 m/s2) = 100 kg. The total oscillating mass is m = 1000 kg + 100 kg = 1100 kg. The period T is:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Vertical SHM### 13.4Applications of SHM

- and the frequency is:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

A simple pendulumis an idealized model consisting of a point mass suspended by a massless, unstretchable string.### 13.5The Simple Pendulum

- When the point mass is pulled to one side of its straight-down equilibrium position and released, it oscillates about the equilibrium position.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The pendulum is a good example of harmonic motion.### 13.5The Simple Pendulum

- Oscillations depend on the length of the pendulum, the gravitational restoring force BUT not the mass.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The path of the point mass (sometimes called a pendulum bob) is not a straight line but the arc of a circle with radius L equal the length of the string (figure above).### 13.5The Simple Pendulum

- If the motion is simple harmonic, the restoring force must be directly proportional to x or (because x = Lθ) to θ. Is it?

- In the figure above we represent the forces on the mass in terms of tangential and radial components.

- The restoring force Fθis the tangential component of the net force:

- The restoring force is provided by gravity; the tension T merely acts to make the point mass move in an arc.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

The restoring force in proportional not to θ but to sin θ, so the motion in not simple harmonic.### 13.5The Simple Pendulum

- If the angle θ is small, sin θ is very nearly equal to θ in radian. For example, when θ = 0.1 rad (about 6o), sin θ = 0.0998.

- A difference of only 0.20%. With this approximation, Eq. (13.30) becomes:

- The restoring force is then proportional to the coordinate for small displacements, and the force constant k = mg/L.

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

From Eq. (13.10) the angular frequency ω of a simple pendulum with small amplitude is:### 13.5The Simple Pendulum

- The corresponding frequency and period relations are:

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

Example 13.8(A simple Pendulum):### 13.5The Simple Pendulum

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

Solution:

- Identify and Set Up:We use Eq. (13.34) to determine the period T of the pendulum from its length, and Eq. (13. 1) to find the frequency f from T.

- Execute:From Eqs. (13.34) and (13.1),

Dr. Y. Abou-Ali, IUST

University Physics, Chapter 13

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