Simple Harmonic

MotionChapter 10Springs

A TRAMPOLINE exerts a restoring force on the jumper that is directly proportional to the average force required to displace the mat. Such restoring forces provide the driving forces necessary for objects that oscillate with simple harmonic motion.

Objectives: After finishing this unit, you should be able to:

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

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

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

Periodic MotionSimple 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.

AmplitudeA

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)

1fT

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 sfT

Frequency: f = 2.00 HzFrequency: f = 2.00 Hz

Hooke’s LawWhen a spring is stretched, there is a restoring force that is proportional to the displacement.

F = -kxThe spring constant k is a property of the spring given by:

k = DFDx

F

x

m

Work Done in Stretching a Spring

F

x

m

Work done ON the spring is positive; work BY spring is negative.To stretch spring from x1 to x2 , work is:

2 22 1½ ½Work kx kx

F (x) = kx

x1 x2

F

Example 2: a) 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 = =

DFDx

39.2 N0.2 m

k = 196 N/mk = 196 N/m

Example 2: b) The mass m is now stretched a distance of 8 cm and held. What is the potential energy? (k = 196 N/m)

F8 cm

m

U = 0.627 JU = 0.627 J

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

2 22 1½ ½Work kx kx 0

2 2½ ½(196 N/m)(0.08 m)U kx PE

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.

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 center of oscillation.

+x-a

-x+a

F ma kx

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

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:

Conservation of EnergyThe total mechanical energy (PE + KE) of a vibrating system is constant; i.e., it is the same at any point in the oscillating path.

m

x = 0 x = +Ax = -A

x va

For any two points A and B, we may write:

½mvA2 + ½kxA 2 = ½mvB

2 + ½kxB 2 ½mvA2 + ½kxA 2 = ½mvB

2 + ½kxB 2

The Period and Frequency as a Function of a and x.

For any body undergoing simple harmonic motion:

1

2

af

x

2x

Ta

The frequency and the period can be found if the displacement and acceleration are known. Note that the signs of a and x will always be opposite.

The frequency and the period can be found if the displacement and acceleration are known. Note that the signs of a and x will always be opposite.

Period and Frequency as a Function of Mass and Spring Constant.

For a vibrating body with an elastic restoring force:

Recall that F = ma = -kx:

1

2

kf

m 2

mT

k

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

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

Example 5: 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?

m

x = 0 x = +0.2 m

x va

x = -0.2 m

1 1 400 N/m

2 2 2 kg

kf

m

f = 2.25 Hzf = 2.25 Hz

Simple Harmonic

MotionChapter 10The Pendulum

A Foucault pendulum consists of a tall pendulum free to oscillate in any vertical plane. The direction along which the pendulum swings rotates with time because of Earth's daily rotation

The Simple PendulumThe period of a simple pendulum is given by:

mg

L

2L

Tg

For small angles .q

1

2

gf

L

Example 6. 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

(2 s) (9.8 m/s )

4L

L = 0.993 m

Chapter 10

Simple Harmonic Motion

1. A) 84,810 N/m B) 288.3 N2. 666.6 N/m3. A) 30 N/m B) 9 N4. 35,092 N/m5. A) 0.45 m B) 3.31 Hz C) 1.49 m/s 6. 1524 N/m 4.29 kg7. A) 46.94 J B) 55.9 m/s8. 1.5 m/s9. 0.556 m/s10. 30 rad/s

Tonight:Pendulum Lab –

write pre-lab

Tomorrow:

Lab Hooke’s Law

Friday:

Pendulum LabFORMAL LAB!!!

18. A person bounces up and down on a trampoline, while always staying in contact with it. The motion is SHM an it takes 1.90 s to complete one cycle. The height of each bounce is 45.0cm. Determine (a) the amplitude (b) the angular frequency (c) What is the maximum speed attained by the person?

Example 9: A man enters a tall tower, needing to know its height h. He notes that a long pendulum extends from the roof almost to the ground and that its period is 15.5 s. (a) How tall is the tower?

a) 59.7 m

(b) If this pendulum is taken to the Moon, where the free-fall acceleration is 1.67 m/s2, what is the period of the pendulum there?

b) 37.6 s