Harmonic Motion

Harmonic MotionChapter 13

Simple Harmonic Motion

A single sequence of moves that constitutes the repeated unit in a periodic motion is called a cycle

The time it takes for a system to complete a cycle is a period (T)

Simple Harmonic MotionThe period is the number of units of time per

cycle; the reciprocal of that—the number of cycles per unit of time—is known as the frequency (f).

The SI unit of frequency is the hertz (Hz), where 1 Hz = 1 cycle/s = 1 s-1

Amplitude (A) is the maximum displacement of an

object in SHM

Tf

1

Simple Harmonic MotionOne complete orbit (one

cycle) object sweeps through 2p rad

f - number of cycles per second

Number of radians it moves through per second is 2pf - that's angular speed ( )w T

f 2

2

w - angular frequency Sinusoidal motion (harmonic) with a single

frequency - known as simple harmonic motion (SHM)

Displacement in SHM

txAx coscos max

Velocity in SHM

tAvx sin

2max )/(1sin AxvtAvx

vmax = Aw

Acceleration in SHM

tAax cos2xax

2

The acceleration of a simple harmonic oscillator is proportional to its displacement

2max Aa

Example 1A spot of light on the screen of a computer

is oscillating to and fro along a horizontal straight line in SHM with a frequency of 1.5 Hz. The total length of the line traversed is 20 cm, and the spot begins the process at the far right. Determine

(a) its angular frequency, (b) its period, (c) the magnitude of its maximum velocity,

and (d) the magnitude of its maximum

acceleration, (e) Write an expression for x and find the

location of the spot at t = 0.40 s.

Example 1Given: f = 1.5 Hz and A = 10 cm Find: (a) w, (b) T, (c) |vx(max)|, (d) |ax(max)|,

(e) x in general, and x at t = 0.40sFrom Eq. (10.12)(a) w = 2pf = 2p(1.5 Hz) = 9.4 rad/s = 3.0p

rad/s(b) T= 1/f= 1/1.5 Hz = 0.67s(c) |vx(max)| = vmax = Aw = 2pfA

vmax = 2p(1.5 Hz)(0.10 m) = 0.94 m/s

Example 1(d) From Eq. (10.17), |ax(max)| = Aw2 = A(2pf)2= (0.10 m)(2p

1.5 Hz)2 = 8.9 m/s2

(e) x = A cos wt = (0.10 m) cos (9.4 rad/s)t At t = 0.40 s x = (0.10m) cos (3.77 rad) = (0.10 m)(-

0.81) = -8.1 cm

Problem

A point at the end of a spoon whose handle is clenched between someone’s teeth vibrates in SHM at 50Hz with an amplitude of 0.50cm. Determine its acceleration at the extremes of each swing.

Equilibrium

The state in which an elastic or oscillating system most wants to be in if undisturbed by outside forces.

Elastic Restoring ForceWhen a system oscillates naturally it moves against a restoring force that returns it to its undisturbed equilibrium condition

A "lossless" single-frequency ideal vibrator is known as a simple harmonic oscillator.

An Oscillating SpringIf a spring with a mass attached

to it is slightly distorted, it will oscillate in a way very closely resembling SHM.

Force exerted by an elastically stretched spring is the elastic restoring force F, = -ks.

Resulting acceleration ax = -(k/m)x

F is linear in x; a is linear in x - hallmark of SHM

Frequency and Period

Simple harmonic oscillator

Shown every ¼ cycles for 2 cycles

Relationship between x, vx, t, and T

Hooke’s Law

Beyond being elastic, many materials deform in proportion to the load they support - Hooke's Law

sF

Hooke’s Law

The spring constant or elastic constant k - a measure of the stiffness of the object being deformed

k has units of N/m

ksF

Hooke’s Law

k has units of N/m

Frequency and Period

w0 - the natural angular frequency, the specific frequency at which a physical system oscillates all by itself once set in motion

natural angular frequency

and since w0 = 2pf0

natural linear frequency

Since T= 1/f0

Period

m

k0

m

kf

21

0

k

mT 2

Resonance vs. Damping• If the frequency of the disturbing force equals the natural frequency of the system, the amplitude of the oscillation will increase—RESONANCE

• If the frequency of the periodic force does NOT equal the natural frequency of the system, the amplitude of the oscillation will decrease--DAMPING

Example 2A 2.0-kg bag of candy is hung on a vertical,

helical, steel spring that elongates 50.0 cm under the load, suspending the bag 1.00 m above the head of an expectant youngster. The candy is pulled down an additional 25.0 cm and released. How long will it take for the bag to return to a height of 1.00 m above the child?

Example 2Given: m = 2.0 kg, ∆L = 50.0 cm, and A =

25.0 cm. Find: t when y = 0.To find T, we must first find kInitially the load (F = Fw) stretches the

springmgLkF )(

mNm

smkg

L

mgk /2.39

50.0

)/81.9)(0.2( 2

smN

kg

k

mT 4.1

/2.39

0.222

¼T = 0.35 s

The PendulumThe period of a

pendulum is independent of the mass and is determined by the square root of its length

L

gf

21

0

g

LT 2

Example 3How long should a pendulum be if it is to

have a period of 1.00 s at a place on Earth where the acceleration due to gravity is 9.81 m/s2?

Given: T = 1.00 s and g = 9.81 m/s2. Find: LFrom T2 = 4π2 (L/g); hence

msmsgT

L 248.04

)/81.9()00.1(

4 2

22

2

2

Problem 2What would the length of a pendulum need to

be on Jupiter in order to keep the same time as a clock on Earth? gJupiter = 25.95m/s2

Tf

1

Tf

22

txAx coscos max

vmax = Aw 2max Aa

ksF m

k0

m

kf

21

0

k

mT 2

When “f” is known

When “f” is NOT known

L

gf

21

0

g

LT 2