Simple Harmonic Motion A trampoline exerts a restoring force on the jumper that is directly proportional to the displacement of the mat. (F s =kx). Such.

Simple Harmonic Motion

A trampoline exerts a restoring force on the jumper that is directly proportional to the displacement of the mat. (Fs=kx). Such restoring forces provide the driving forces necessary for objects that oscillate with simple harmonic motion.

Simple Harmonic MotionSimple harmonic motion Back and forth motion that is caused by a force that is directly proportional to the displacement. (F=kx) The displacement centers around an equilibrium position.

AmplitudeA

Period, T, is the time for one complete oscillation. (seconds)

Period, T, is the time for one complete oscillation. (seconds)

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)

1fT

Example 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 cyclesT

Period: T = 0.5 sPeriod: T = 0.5 s

1 1

0.5 sfT

Frequency: f = 2 HzFrequency: f = 2 Hz

ExampleA load of 50 N attached to a spring hanging vertically stretches the spring 5.0 cm. The spring is now placed horizontally on a table and stretched 11.0 cm. What force is required to stretch the spring this amount?

k

k

kxFs)05.0(50

1000 N/m

s

s

s

F

F

kxF

)11.0)(1000(

110 N

Position graph of a spring

When you graph the motion of a spring, it is a sine function

Position time graph for SHMThe amplitude of the curve is the max displacement ,x

Equilibrium Line

Period, T, is the time for one revolution or in the case of springs the time for ONE COMPLETE oscillation (One crest and trough). Oscillations could also be called vibrations and cycles. In the wave above we have 1.75 cycles or waves or vibrations or oscillations.

SHM and Uniform Circular MotionSprings (and waves) behave very similar to objects that move in

circles.

If you trace out the position of an object on the outside of a circle, you get a sine function (remember the unit circle??)

circlewavespring rAx

http://www.animations.physics.unsw.edu.au//jw/SHM.htm#projection

Period of a spring

For a vibrating body with an elastic restoring force:

Recall that F = ma = kx:

2m

Tk

The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known.

The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known.

2

2

2

2 2

22

2

4

4

4

F ma kx

rm

Tkx

r

m rkr

T

T kr m r

mT

k

Example

A 200 g mass is attached to a spring and executes simple harmonic motion with a period of 0.25 s If the total energy of the system is 2.0 J, find the (a) force constant of the spring (b) the amplitude of the motion

kkk

mTs

200.0225.02

2 21 122 2SPE kx kA A

126.3 N/m

0.18 m

Velocity in SHM

m

x = 0 x = +Ax = -A

v (+)

• Velocity is positive when moving to the right and negative when moving to the left.• It is zero at the end points and a maximum at the midpoint in either direction (+ or -).

v (-)

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 midpoint of oscillation.

+x-a

-x+a

Example 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

Example: 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:

Graph of Force vs. displacement

Remember that the slope of a force vs. displacement graph is equal to k

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0

10

20

30

40

50

60

70

80

f(x) = 120 xR² = 1

Force vs. Displacement

Displacement(Meters)

Fo

rce

(Ne

wto

ns

)

k =120 N/m

Graph of Force vs. displacement

And that the area is equal to the spring potential energy 21

2SPE kx

ExampleA slingshot consists of a light leather cup, containing a stone, that

is pulled back against 2 rubber bands. It takes a force of 30 N to stretch the bands 1.0 cm (a) What is the potential energy stored in the bands when a 50.0 g stone is placed in the cup and pulled back 0.20 m from the equilibrium position? (b) With what speed does it leave the slingshot?

) 30 (0.01)sa F kx k k 3000 N/m

60 J

49 m/s

2 21 0.5( )(.20)2SPE kx k

2 2

)

1 1 (0.050)2 2

B Ab E E SPE KE

SPE mv v

v

PendulumsPendulums, like springs, oscillate

back and forth exhibiting simple harmonic behavior.

http://www.walter-fendt.de/ph14e/pendulum.htm

The restoring force is the force that brings the pendulum back to its equilibrium

PendulumsConsider the FBD for a pendulum. Here we have the weight and tension. Even though the weight isn’t at an angle let’s draw an axis along the tension.

q

q

mgcosq

mgsinq

kxmg

mg

sin

Force Restoringsin

Pendulumskxmg

mg

sin

Force Restoringsin

k

mT

g

l

k

m

klmg

smallif

Lkmg

AmplitudeLsL

s

R

s

spring

2

,sin

sin

What is x? It is the amplitude! In the picture to the right, it represents the chord from where it was released to the bottom of the swing (equilibrium position).

g

lTpendulum 2

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

2L

Tg

L

22 2

24 ; L =

4

L T gT

g

2

2

(2) (10)

4L

L = 1.01 m

ExampleA visitor to a lighthouse wishes to determine the

height of the tower. She ties a spool of thread to a small rock to make a simple pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 9.40 s. What is the height of the tower?

22 22

2 2

2

4 9.4 (10)

4 4( )

P

PP

lT l height

g

T glT l

g

L = Height = 22.4 m