Physics 207: Lecture 19, Pg 1 Lecture 19Goals: Chapter 14 Chapter 14 Periodic motion. Assignment Assignment No HW this week. Wednesday: Read through

Physics 207: Lecture 19, Pg 1

Lecture 19Goals:Goals:

• Chapter 14Chapter 14 Periodic motion.

• AssignmentAssignment No HW this week. Wednesday: Read through Chapter 15.4


Periodic Motion is everywhere

Examples of periodic motion Earth around the sun Elastic ball bouncing up and down Quartz crystal in your watch, computer clock, iPod clock, etc.


Periodic Motion is everywhere

Examples of periodic motion Heart beat

In taking your pulse, you count 70.0 heartbeats in 1 min.

What is the period, in seconds, of your heart's oscillations?

Period is the time for one oscillation

T= 60 sec/ 70.0 = 0.86 s


Simple Harmonic Motion (SHM)

We know that if we stretch a spring with a mass on the end and let it go the mass will, if there is no friction, ….do something

1. Pull block to the right until x = A

2. After the block is released from x = A, it will

A: remain at rest

B: move to the left until it reaches

equilibrium and stop there

C: move to the left until it reaches

x = -A and stop there

D: move to the left until it reaches

x = -A and then begin to move to

the right

km

-A A0(≡Xeq)



The time it takes the block to complete one cycle is the period T and is measured in seconds.

The frequency, denoted f, is the number of cycles that are completed per unit of time: f = 1 / T.

In SI units, f is measured in inverse seconds, or hertz (Hz).

If the period is doubled, the frequency is

A. unchanged

B. doubled

C. halved



An oscillating object takes 0.10 s to complete one cycle; that is, its period is 0.10 s.

What is its frequency f ?

Express your answer in hertz.

f = 1/ T = 10 Hz


Simple Harmonic Motion Suppose that the period is T. Which of the following points on the t axis are separated by the

time interval T?

A. K and L

B. K and M

C. K and P

D. L and N

E. M and P

time


Simple Harmonic Motion Now assume that the t coordinate of point K is 0.25 s. What is the period T , in seconds?

How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement?

time



T = 1 s How much time t does the block take to travel from the point of

maximum displacement to the opposite point of maximum displacement?

time



T = 1 s How much time t does the block take to travel from the point of

maximum displacement to the opposite point of maximum displacement?

t = 0.5 s

time


Simple Harmonic Motion

Now assume that the x coordinate of point R is 1 m. What total distance d does the object cover during one period of

oscillation?

What distance d does the object cover between the moments

labeled K and N on the graph?

time




oscillation?

d = 4 m What distance d does the object cover between the moments


time




oscillation?

d = 4 m What distance d does the object cover between the moments


d = 3 m

time


km

-A A0(≡Xeq)

T = 1 s

k2m

-A A0(≡Xeq)

T is:

A)T > 1 s

B)T < 1 s

C) T=1 s


km

-A A0(≡Xeq)

T = 1 s

2km

-A A0(≡Xeq)

T is:

A)T > 1 s

B)T < 1 s

C)T=1 s


SHM Dynamics: Newton’s Laws still apply

At any given instant we know that F = ma must be true.

But in this case F = -k x and ma = F

So: -k x = ma =m d2x/dt2

k

x

m

FF = -k x

aa

a differential equation for x(t) !

2

2

dt

xdm

“Simple approach”, guess a solution and see if it works!

d2x/dt2=-(k/m)x


SHM Solution... Try cos ( t ) Below is a drawing of A cos ( t ) where A = amplitude of oscillation

[with = (k/m)½ and = 2 f = 2 /T ] T=2 (m/k)½

T = 2/

A

A


Use “initial conditions” to determine phase !

The general solution is: x(t) = A cos ( t + )

SHM Solution...

km

-A A0(≡Xeq)

x(t) = A cos ( t + )

km

-A A0(≡Xeq)

x(t) = A cos ( t + )

at t=0

at t=0


Energy of the Spring-Mass System

We know enough to discuss the mechanical energy of the oscillating mass on a spring.

x(t) = A cos ( t + ) If x(t) is displacement from equilibrium, then potential energy is

U(t) = ½ k x(t)2 = ½ k A2 cos2 ( t + )

v(t) = dx/dt v(t) = A -sin ( t + )) And so the kinetic energy is just ½ m v(t)2

K(t) = ½ m v(t)2 = ½ m (A)2 sin2 ( t + ) Finally,

a(t) = dv/dt = -2A cos(t + )



Potential energy of the spring is

U = ½ k x2 = ½ k A2 cos2(t + )

The Kinetic energy is

K = ½ mv2 = ½ m(A)2 sin2(t+)

And 2 = k / m or k = m 2

K = ½ k A2 sin2(t+)

x(t) = A cos( t + )

v(t) = -A sin( t + )

a(t) = -2A cos( t + )



So E = K + U = constant =½ k A2

m

k

m

k 2

U~cos2K~sin2

E = ½ kA2

At maximum displacement K = 0 and U = ½ k A2

and acceleration has it maximum

At the equilibrium position K = ½ k A2 = ½ m v2 and U = 0


SHM So Far

The most general solution is x = A cos(t + )

where A = amplitude

= (angular) frequency

= phase constant

For SHM without friction,

The frequency does not depend on the amplitude ! This is true of all simple harmonic motion!

The oscillation occurs around the equilibrium point where the force is zero!

Energy is a constant, it transfers between potential and kinetic

mk


The “Simple” Pendulum

A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements.

Fy = may = T – mg cos() ≈ m v2/L

Fx = max = -mg sin()

If small then x L and sin() dx/dt = L d/dt

ax = d2x/dt2 = L d2/dt2

so ax = -g = L d2/ dt2

and = cos(t + )

with = (g/L)½

L

m

mg

z

y

x

T


The shaker cart You stand inside a small cart attached to a heavy-duty spring, the spring

is compressed and released, and you shake back and forth, attempting to maintain your balance. Note that there is also a sandbag in the cart with you.

At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground.

What effect does jettisoning the sandbag at the equilibrium position have on the amplitude of your oscillation?

It increases the amplitude.It decreases the amplitude.It has no effect on the amplitude.

Hint: At equilibrium, both the cart and the bag are moving at their maximum speed. By dropping the bag at this point, energy (specifically the kinetic energy of the bag) is lost from the spring-cart system. Thus, both the elastic potential energy at maximum displacement and the kinetic energy at equilibrium must decrease


The shaker cart Instead of dropping the sandbag as you pass through equilibrium, you

decide to drop the sandbag when the cart is at its maximum distance from equilibrium.

What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the amplitude of your oscillation?

It increases the amplitude.

It decreases the amplitude.

It has no effect on the amplitude. Hint: Dropping the bag at maximum

distance from equilibrium, both the cart

and the bag are at rest. By dropping the

bag at this point, no energy is lost from

the spring-cart system. Therefore, both the

elastic potential energy at maximum displacement

and the kinetic energy at equilibrium must remain constant.


The shaker cart

What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the maximum speed of the cart?

It increases the maximum speed.

It decreases the maximum speed.

It has no effect on the maximum speed.

Hint: Dropping the bag at maximum distance

from equilibrium, both the cart and the bag

are at rest. By dropping the bag at this

point, no energy is lost from the spring-cart

system. Therefore, both the elastic

potential energy at maximum displacement

and the kinetic energy at equilibrium must

remain constant.

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Physics 207: Lecture 19, Pg 1 Lecture 19Goals: Chapter 14 Chapter 14 Periodic motion. Assignment Assignment No HW this week. Wednesday: Read through