Simple Harmonic Motion

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# Simple Harmonic Motion - PowerPoint PPT Presentation

Simple Harmonic Motion. An Introduction. Periodic Motion. Repeated motion along the same path over a fixed reproducible period of time Acrobat on a trapeze Wrecking ball Playground swing Mass attached to a spring. x. m. L. x = -A. x = 0. x = +A. Oscillators.

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### Simple Harmonic Motion

An Introduction

Periodic Motion

Repeated motion along the same path over a fixed reproducible period of time

• Acrobat on a trapeze
• Wrecking ball
• Playground swing
• Mass attached to a spring

x

m

L

x = -A

x = 0

x = +A

Oscillators
• Mechanical devices that create this motion
• Weight attached to spring vertically
• Weight attached to spring horizontally
• Pendulum
If you were to plot distance vs time you would get a graph that resembled a sine or cosine function.

3

t(s)

2

4

6

-3

x(m)

Simple Harmonic Motion (SHM)
• Periodic motion which can be described by a sine or cosine function.
• Springs and pendulums are common examples of Simple Harmonic Oscillators (SHOs).
Equilibrium
• The midpoint of the oscillation of a simple harmonic oscillator.
• Position of minimum potential energy and maximum kinetic energy.
Ug increases as pendulum’s displacement increases

Maximum

Displacement

Maximum

Displacement

Equilibrium

Us increases as pendulum’s displacement increases

Maximum

Displacement

Maximum

Displacement

Equilibrium

m

m

m

All oscillators obey…

Law of Conservation of Energy

Amplitude A

Amplitude (A)
• How far the body is from equilibrium at its maximum displacement.
• Higher amplitude equals more energy
Period (T)
• The length of time it takes for one cycle of periodic motion to complete itself.
Frequency (f):
• How fast the oscillation is occurring.
• Frequency is inversely related to period.
• f = 1/T
• The units of frequency is the Hertz (Hz) where 1 Hz = 1 s-1.

x

F

Example 1:The suspended mass makes 30 complete oscillations in 15 s. What is the period and frequency of the motion?

Period: T = 0.500 s

Frequency: f = 2.00 Hz

T

A

Equilibrium point

Parts of a Wave

3

t(s)

2

4

6

-3

x(m)

Springs
• A very common type of Simple Harmonic Oscillator.
• Our springs will be ideal springs.
• They are massless.
• They are compressible and extensible.
• They are restored to equilibrium according to Hooke’s Law: Fs=-kx
Restoring force
• The restoring force is the secret behind simple harmonic motion.
• The force is always directed so as to push or pull the system back to its equilibrium (normal rest) position.
Hooke’s Law

A restoring force directly proportional to displacement is responsible for the motion of a spring.

F = -kx

where

F: restoring force

k: force constant

x: displacement from equilibrium

Fs

m

mg

Hooke’s Law

The force constant of a spring can be determined by attaching a weight and seeing how far it stretches.

Fs = -kx

x

m

F

m

F

x

m

Hooke’s Law

F = -kx

Equilibrium position

Spring compressed, restoring force out

Spring at equilibrium, restoring force zero

Spring stretched, restoring force in

0.20m

F

m

DF

Dx

39.2 N

0.2 m

k = =

A 4-kg mass suspended from a spring produces a displacement of 20 cm. What is the spring constant?

The stretching force is the weight (W = mg) of the 4-kg mass:

F = (4 kg)(9.8 m/s2) = 39.2 N

Now, from Hooke’s law, the force constant k of the spring is:

k = 196 N/m

0.08 m

F

0

m

The mass m is now stretched a distance of 8 cm and held. What is the potential energy? (k = 196 N/m)

The potential energy is equal to the work done in stretching the spring:

U = 0.627 J

+a

-a

-x

+x

m

x = -A

x = 0

x = +A

Acceleration in SHM
• Acceleration is in the direction of the restoring force. (a ispositive when x is negative, and negative when x is positive.)
• Acceleration is a maximum at the end points and it is zero at the center of oscillation.

a

v

x

m

x = -A

x = 0

x = +A

Acceleration vs. Displacement

Given the spring constant, the displacement, and the mass, the acceleration can be found from:

or

Note: Acceleration is always oppositeto displacement.

a

+x

m

A 2-kg mass hangs at the end of a spring whose constant is k = 400 N/m. The mass is displaced a distance of 12 cm and released. What is the acceleration at the instant the displacement is x = +7 cm?

a = -14.0 m/s2

Note: When the displacement is +7 cm (downward), the acceleration is -14.0 m/s2 (upward) independent of motion direction.

+x

m

What is the maximumacceleration for the 2-kg mass in the previous problem? (A = 12 cm, k = 400 N/m)

The maximum acceleration occurs when the restoring force is a maximum; i.e., when the stretch or compression of the spring is largest.

xmax=  A

F = ma = -kx

Maximum Acceleration:

amax = ± 24.0 m/s2

Period of a spring
• T = 2m/k
• T: period (s)
• m: mass (kg)
• k: force constant (N/m)

m

a

v

x

x = 0

x = -0.2 m

x = +0.2 m

The frictionless system shown below has a 2-kg mass attached to a spring (k = 400 N/m). The mass is displaced a distance of 20 cm to the right and released.What is the frequency of the motion?

f = 2.25 Hz

Pendulums
• Springs are not the only type of oscillators.
• The pendulum can be thought of as an oscillator.
What factor(s) determine the period of a pendulum??A) MassB) Length of StringC) GravityD) Angle of displacement

L

T

mg

The Simple Pendulum

The period of a simple pendulum is given by:

For small angles q.

L

L = 0.993 m

What must be the length of a simple pendulum for a clock which has a period of two seconds (tick-tock)?