Unit 2:

Simple Harmonic Motion

(SHM)

THE MOST COMMON FORM OF MOTION

FALL 2015

Objectives:

• Define SHM specifically and give an example.

• Describe the motion of pendulums and spring systems and calculate the length required to produce a given frequency.

• Write and apply Hooke’s Law for objects moving with simple harmonic motion.

• Write and apply formulas for finding the frequency f, period T, w angular frequency, velocity v, or acceleration a in terms of displacement x or time t.

Displacement in SHM

m

x = 0 x = +Ax = -A

x

• Displacement is positive when the position is

to the right of the equilibrium position (x = 0)

and negative when located to the left.

• The maximum displacement is called the

amplitude A.

Periodic Motion

Periodic motion is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time.

Amplitude

A

Period, T, is the time

for one complete

oscillation. (seconds,s)

Period, T, is the time

for one complete

oscillation. (seconds,s)

Frequency, f, is the

number of complete

oscillations per

second. Hertz (s-1)

Frequency, f, is the

number of complete

oscillations per

second. Hertz (s-1)

1f

T

Example 1: The suspended mass makes

30 complete oscillations in 15 s. What is

the period and frequency of the

motion?

x F

15 s0.50 s

30 cylcesT

Period: T = 0.500 sPeriod: T = 0.500 s

1 1

0.500 sf

T

Frequency: f = 2.00 HzFrequency: f = 2.00 Hz

Simple Harmonic Motion,

SHMSimple harmonic motion:

• periodic motion in the absence of friction

(Amplitude is always the same/ Energy is

Conserved)

• produced by a restoring Force that is

directly proportional to the displacement

and oppositely directed (restoring to

equilibrium).

• Displacement does not affect the frequency/

The energy of the system does not affect the

frequency.

7

https://www.youtube.com/watch?v=SZ

541Luq4nE

https://www.youtube.com/watch?

v=VnGkoMoUkgI

https://www.youtube.com

/watch?v=eeYRkW8V7Vg

Conceptual Example 4. Moving Lights10

Over the entrance to a restaurant is mounted a strip of equally spaced

light bulbs, as Figure 10.13a illustrates. Starting at the left end, each

bulb turns on in sequence for one-half second. Thus, a lighted bulb

appears to move from left to right.

Once the apparent motion of a

lighted bulb reaches the right side

of the sign, the motion reverses.

The lighted bulb then appears to

move to the left, as part b of the

drawing indicates. Thus, the

lighted bulb appears to oscillate

back and forth. Is the apparent

motion simple harmonic motion?

No. Speed is constant.

Damped Harmonic Motion

13

In the presence of energy dissipation, the amplitude of

oscillation decreases as time passes, and the motion is no

longer simple harmonic motion. Instead, it is referred to as

damped harmonic motion, the decrease in amplitude being

called “damping.”

14

The smallest degree of damping that completely eliminates the

oscillations is termed “critical damping,” and the motion is said to

be critically damped.

When the damping exceeds the critical value, the motion is said to

be overdamped. In contrast, when the damping is less than the

critical level, the motion is said to be underdamped (curves 2 and

3).

16

The maximum excursion from equilibrium is the amplitude A

of the motion. The shape of this graph is characteristic of

simple harmonic motion and is called “sinusoidal,” because it

has the shape of a trigonometric sine or cosine function.

Simple Harmonic Motion and

the Reference Circle

17

Simple harmonic motion, like any motion, can be described

in terms of displacement, velocity, and acceleration.

18DISPLACEMENT

19For any object in simple

harmonic motion, the time

required to complete one

cycle is the period, T (MKS)

Instead of the period, it is more

convenient to speak of the angular

frequency w of the motion, the

frequency being just the number of

cycles of the motion per second.

20

One cycle per second is referred to as one hertz (Hz).

One thousand cycles per second is called one kilohertz

(kHz).

is often called the angular frequency.w

21FREQUENCY OF VIBRATION

F = ma

tAmtAk www coscos 2

sradinm

k/ww

Body Mass Measurement Device

Astronauts who spend long periods of time in orbit periodically

measure their body masses as part of their health-maintenance

programs. On earth, it is simple to measure body weight W with a

scale and convert it to mass m using the acceleration due to gravity,

since W = mg. However, this procedure does not work in orbit, because

both the scale and the astronaut are in free-fall and cannot press

against each other. Instead, astronauts use a body mass measurement

device. This device consists of a spring-mounted chair in which the

astronaut sits. The chair is then started oscillating in simple harmonic

motion. The period of the motion is measured electronically and is

automatically converted into a value of the astronaut’s mass, after the

mass of the chair is taken into account. The spring used in one such

device has a spring constant of 606 N/m, and the mass of the chair is

12.0 kg. The measured oscillation period is 2.41 s. Find the mass of the

astronaut.

23

m

k

T

w

2

Velocity in SHM

m

x = 0 x = +Ax = -A

v (+)

• Velocity is positive when moving to the rightand negative when moving to the left.

• It is zero at the end points and a maximumat the midpoint in either direction (+ or -).

v (-)

Velocity as Function of

Position.

m

x = 0 x = +Ax = -A

x va

kv A

m

vmax when

x = 0:

2 2kv A x

m 2 2 21 1 1

2 2 2mv kx kA

Check Your Understanding 2 26

The drawing shows plots

of the displacement x

versus the time t for three

objects undergoing simple

harmonic motion. Which

object, I, II, or III, has the

greatest maximum

velocity?

II

Example 5: A 2-kg mass hangs at the end

of a spring whose constant is k = 800 N/m.

The mass is displaced a distance of 10 cm

and released. What is the velocity at the

instant the displacement is x = +6 cm?

m+x

½mv2 + ½kx 2 = ½kA2

2 2kv A x

m

2 2800 N/m(0.1 m) (0.06 m)

2 kgv

v = ±1.60 m/sv = ±1.60 m/s

Example 5 (Cont.): What is the maximum

velocity for the previous problem? (A = 10

cm, k = 800 N/m, m = 2 kg.)

m+x

½mv2 + ½kx 2 = ½kA2

800 N/m(0.1 m)

2 kg

kv A

m

v = ± 2.00 m/sv = ± 2.00 m/s

0

The velocity is maximum when x = 0:

Example 3. The Maximum

Speed of a Loudspeaker

Diaphragm

29

The diaphragm of a loudspeaker moves back and forth in

simple harmonic motion to create sound. The frequency of the

motion is f = 1.0 kHz and the amplitude is A = 0.20 mm.

(a)What is the maximum

speed of the diaphragm?

(b)Where in the motion

does this maximum speed

occur?

30

(b) The speed of the diaphragm is zero when the

diaphragm momentarily comes to rest at either end of its

motion: x = +A and x = –A. Its maximum speed occurs

midway between these two positions, or at x = 0 m.

(a)

Acceleration in SHM

m

x = 0 x = +Ax = -A

• Acceleration is in the direction of the restoring force. (a is positive 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.

+x-a

-x+a

F ma kx F ma kx

Acceleration vs.

Displacement

m

x = 0 x = +Ax = -A

xv

a

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

or

Note: Acceleration is always opposite to displacement.

F ma kx F ma kx kx

am

kxa

m

Example 3: 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?

m+x

(400 N/m)(+0.07 m)

2 kga

a = -14.0 m/s2a = -14.0 m/s2

a

Note: When the displacement is +7 cm(downward), the acceleration is -14.0 m/s2

(upward) independent of motion direction.

kxa

m

kxa

m

Example 4: What is the maximum

acceleration for the 2-kg mass in

the previous problem? (A = 12 cm,

k = 400 N/m)

m+x

The maximum acceleration occurs when the restoring force is a maximum; i.e., when the stretch or compression of the spring is largest.F = ma = -kx xmax = A

400 N( 0.12 m)

2 kg

kAa

m

amax = ± 24.0 m/s2amax = ± 24.0 m/s2Maximum

Acceleration:

Example 5.

The Loudspeaker Revisited—The

Maximum Acceleration

35

A loudspeaker diaphragm is

vibrating at a frequency of

f = 1.0 kHz, and the

amplitude of the motion is

A = 0.20 mm.

(a)What is the maximum

acceleration of the

diaphragm, and

(b)where does this maximum

acceleration occur?

36

(b) the maximum acceleration occurs at x = +A and x = –A

(a)

Hooke’s Law

When a spring is stretched, there is a restoringforce that is proportional to the displacement.

F = -kx

The spring constant k is a property of the spring given by:

k = DF

Dx

F

x

m

Ch10. Simple Harmonic Motion and

Elasticity

kxFApplied

kxFApplied

38

The Ideal Spring and Simple Harmonic Motion

The constant k is called the

spring constant

A spring that behaves

according to is

said to be an ideal spring.

39The restoring force also

leads to simple harmonic

motion when the object is

attached to a vertical

spring, just as it does

when the spring is

horizontal. When the

spring is vertical,

however, the weight of

the object causes the

spring to stretch, and the

motion occurs with

respect to the

equilibrium position of

the object on the

stretched spring .mg = kd0, which gives d0 = mg/k.

Example 1. A Tire Pressure Gauge

40

In a tire pressure gauge, the air in

the tire pushes against a plunger

attached to a spring when the

gauge is pressed against the tire

valve. Suppose the spring

constant of the spring is k = 320

N/m and the bar indicator of the

gauge extends 2.0 cm when the

gauge is pressed against the tire

valve. What force does the air in

the tire apply to the spring?

Conceptual Example 2.

Are Shorter Springs Stiffer Springs?41

A 10-coil spring that has a spring

constant k. If this spring is cut in

half, so there are two 5-coil

springs, what is the spring

constant of each of the smaller

springs?

Shorter springs are stiffer springs.

Sometimes the spring constant k is referred to as

the stiffness of the spring, because a large value

for k means the spring is “stiff,” in the sense that

a large force is required to stretch or compress it.

Example 2: A 4-kg mass suspended from a

spring produces a displacement of 20 cm.

What is the spring constant?

F20 cm

m

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 = =DF

Dx

39.2 N

0.2 mk = 196 N/mk = 196 N/m

Check Your Understanding 1 43

A 0.42-kg block is attached to the end of a horizontal ideal

spring and rests on a frictionless surface. The block is pulled

so that the spring stretches by 2.1 cm relative to its

unstrained length. When the block is released, it moves with

an acceleration of 9.0 m/s2. What is the spring constant of

the spring?

180 N/m

44

2.1c

mkx = ma

2/0.942.0100

1.2smk

mNk /1801001.2

0.942.0