Physics 1501: Lecture 25 Today ’ s Agenda

Physics 1501: Lecture 25, Pg 1

Physics 1501: Lecture 25Physics 1501: Lecture 25TodayToday’’s Agendas Agenda

Homework #9 (due Friday Nov. 4)

Midterm 2: Nov. 16

TopicsReview of static equilibriumOscillationSimple Harmonic Motion – masses on springs Energy of the SHO


Approach to Statics:Approach to Statics:

In general, we can use the two equations

to solve any statics problems.

When choosing axes about which to calculate torque, we can be clever and make the problem easy....


Lecture 25, Lecture 25, Act 1Act 1StaticsStatics

A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a white dot in each case.In which cases does the box tip over ?

(a) all (b) 2 & 3 (c) 3 only

1 2 3


Lecture 25, Lecture 25, Act 1Act 1 SolutionSolution

We have seen that the torque due to gravity acts as though all the mass of an object is concentrated at the center of mass.

Consider the bottom right corner of the box to be a pivot point.

If the box can rotate in such a way that the center of mass islowered, it will !


Lecture 25, Lecture 25, ACT 1ACT 1 SolutionSolution

We have seen that the torque due to gravity acts as though all the mass of an object is concentrated at the center of mass.

Consider the bottom right corner of the box to be a pivot point.

If the box can rotate in such a way that the center of mass islowered, it will !


Lecture 25, Lecture 25, Act 1Act 1 AddendumAddendum

What are the torques ??(where do the forces act ?)

rG

mg g

switches sign at critical point

rf

f

f

always zero

goes to zero at critical point

rN

N

N


New topic (Ch. 13) New topic (Ch. 13) Simple Harmonic Motion (SHM)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 oscillate back and forth (if there is no friction).

This oscillation is called Simple Harmonic Motion,and is actually very easy to understand...

km

km

km


SHM DynamicsSHM Dynamics

At any given instant we know that FF = maa must be true.

But in this case F = -kx

and ma =

So: -kx = ma =

k

x

m

FF = -kx

aa

a differential equation for x(t) !


SHM Dynamics...SHM Dynamics...

Try the solution x = Acos(t)

this works, so it must be a solution !

define


SHM SolutionSHM Solution

We just showed that (which came from F=ma)

has the solution x = Acos(t) .

This is not a unique solution, though. x = Asin(t) is also a solution.

The most general solution is a linear combination of these two solutions!

x = Bsin(t)+ Ccos(t)

ok


Derivation:Derivation:

x = Acos(t+) is equivalent to x = Bsin(t)+ Ccos(t)

x = Acos(t+)

= Acos(t) cos - Asin(t) sin

where C = Acos() and B = Asin()

It works!

= Ccos(t) + Bsin(t)

We want to use the most general solution:

So we can use x = Acos(t+) as the most general solution!


SHM Solution...SHM Solution...

Drawing of Acos(t ) A = amplitude of oscillation

T = 2/

A

A



Drawing of Acos(t + )



Drawing of Acos(t - /2)

A

= Asin(t) !


What about Vertical Springs?What about Vertical Springs?

We already know that for a vertical spring

if y is measured from the equilibrium position

The force of the spring is the negative derivative of this function:

So this will be just like the horizontal case:

-ky = ma =

j j

k

mF= -ky

y = 0

Which has solution y = Acos(t + )

where


Velocity and AccelerationVelocity and Acceleration

k

x

m

0

Position: x(t) = Acos(t + )

Velocity: v(t) = -Asin(t + )

Acceleration: a(t) = -2Acos(t + )

by taking

derivatives,

since:

xMAX = AvMAX = AaMAX = 2A


Lecture 25, Lecture 25, Act 2Act 2Simple Harmonic MotionSimple Harmonic Motion

A mass oscillates up & down on a spring. It’s position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration ?

t

y(t)

(a)

(b)

(c)


ExampleExample A mass m = 2kg on a spring oscillates with amplitude A =

10cm. At t=0 its speed is maximum, and is v = +2 m/s.What is the angular frequency of oscillation ?What is the spring constant k ?

k

x

m

=

Also: k = m2

So k = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m

vMAX = A


Initial ConditionsInitial Conditions

k

x

m

0

Use “initial conditions” to determine phase !

Suppose we are told x(0) = 0 , and x is

initially increasing (i.e. v(0) = positive):

x(t) = Acos(t + )

v(t) = -Asin(t + )

a(t) = -2Acos(t + )

sincos

x(0) = 0 = Acos() = /2 or -/2

v(0) > 0 = -Asin() < 0

= -/2So


Initial Conditions...Initial Conditions...

k

x

m

0

x(t) = Acos(t - /2 )

v(t) = -Asin(t - /2 )

a(t) = -2Acos(t - /2 )

So we find = -/2 !!

x(t) = Asin(t)

v(t) = Acos(t)

a(t) = -2Asin(t)

t

x(t)A

-A


Lecture 25, Lecture 25, Act 3Act 3Initial ConditionsInitial Conditions

A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describe its velocity and acceleration as a function of time (upwards is positive y direction):

k

m

y

0

d

(a) v(t) = - vmax sin(t) a(t) = -amax cos(t)

(b) v(t) = vmax sin(t) a(t) = amax cos(t)

(c) v(t) = vmax cos(t) a(t) = -amax cos(t)

(both vmax and amax are positive numbers)

t = 0


Energy of the Spring-Mass SystemEnergy of the Spring-Mass System

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

Kinetic energy is always K = 1/2 mv2

K = 1/2 m (-Asin(t + ))2

We also know what the potential energy of a spring is,

U = 1/2 k x2

U = 1/2 k (Acos(t + ))2

x(t) = Acos(t + )

v(t) = -Asin(t + )

a(t) = -2Acos(t + )

Remember,


Energy of the Spring-Mass SystemEnergy of the Spring-Mass System

Add to get E = K + U

1/2 m (A)2sin2(t + ) + 1/2 k (Acos(t + ))2

Remember that

U~cos2K~sin2

E = 1/2 kA2

so, E = 1/2 kA2 sin2(t + ) + 1/2 kA2 cos2(t + )

= 1/2 kA2 [ sin2(t + ) + cos2(t + )]

= 1/2 kA2


SHM So FarSHM So Far

The most general solution is x = Acos(t + )

where A = amplitude

= frequency

= phase constant

For a mass on a spring

The frequency does not depend on the amplitude !!!We will see that this is true of all simple harmonic

motion ! The oscillation occurs around the equilibrium point where

the force is zero!


The Simple PendulumThe 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.

L

m

mg

z


The Simple Pendulum...The Simple Pendulum...

Recall that the torque due to gravity about the rotation (z) axis is = -mgd.

d = Lsin Lfor small

so = -mg L

L

dm

mg

z

where

Differential equation for simple harmonic motion !

= 0 cos(t + )

But = II=mL2


Lecture 25, Lecture 25, Act 4Act 4Simple Harmonic MotionSimple Harmonic Motion

You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1.

Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2.

Which of the following is true:

(a) T1 = T2

(b) T1 > T2

(c) T1 < T2


The Rod PendulumThe Rod Pendulum

A pendulum is made by suspending a thin rod of length L and mass M at one end. Find the frequency of oscillation for small displacements.

Lmg

z

xCM


The Rod Pendulum...The Rod Pendulum...

The torque about the rotation (z) axis is

= -mgd = -mg{L/2}sin-mg{L/2}for small

In this case

Ldmg

z

L/2

xCM

where

d I

So = Ibecomes


Lecture 25, Lecture 25, Act 25Act 25PeriodPeriod

(a) (b) (c)

What length do we make the simple pendulum so that it has the same period as the rod pendulum?

LR

LS

Physics 1501: Lecture 25 Today ’ s Agenda

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