Lect 25 Fisika Dasar

Physics 111: Lecture 25 Today’s Agenda ● ● ● ● ●

● ● ●

Recap of last lecture Using “initial conditions” to solve problems The general physical pendulum The torsion pendulum Energy in SHM ➧ Atomic Vibrations Problem: Vertical Spring Problem: Transport Tunnel SHM Review

Physics 111: Lecture 25, Pg 1

SHM and Springs

Force:

d 2s 2 = −ω s 2 dt

ω=

k m

k s Solution: s = A cos(ωt + φ)

k

m

0

0 m

s Physics 111: Lecture 25, Pg 2

Velocity and Acceleration x(t) = A cos(ωt + φ) v(t) = -ωA sin(ωt + φ) a(t) = -ω2A cos(ωt + φ)

Position: Velocity: Acceleration:

xMAX = A vMAX = ωA aMAX = ω2A k

m

0

by taking derivatives, since: v( t ) =

dx ( t ) dt

a( t ) =

dv ( t ) dt

x Physics 111: Lecture 25, Pg 3

Lecture 25, Act 1 Simple Harmonic Motion ●

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

y(t) (a) (c) t (b)

Physics 111: Lecture 25, Pg 4

Lecture 25, Act 1 Solution ●

The slope of y(t) tells us the sign of the velocity since v =

●

y(t) and a(t) have the opposite sign since a(t) = -ω2 y(t)

a<0 v<0

y(t)

dy dt

a<0 v>0

(a) (c) t (b) a>0 v>0

The answer is (c). Physics 111: Lecture 25, Pg 5

Example ●

A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm. At t = 0 its speed is maximum, and is v = +2 m/s. ➧ What is the angular frequency of oscillation ω? ➧ What is the spring constant k? v MAX 2 m s = = 20 s −1 vMAX = ωA ω= A 10 cm k Also: ω = k = mω2 m So k = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m k

m x Physics 111: Lecture 25, Pg 6

Initial Conditions Use “initial conditions” to determine phase φ! x(t) = A cos(ωt + φ) v(t) = -ωA sin(ωt + φ) Suppose we are told x(0) = 0 , and x is 2 a(t) = -ω A cos(ωt + φ) initially increasing (i.e. v(0) = positive): φ = π/2 or -π/2 φ<0

x(0) = 0 = A cos(φ) v(0) > 0 = -ωA sin(φ) So φ = -π/2

π k

cos sin

m

0

2π

x Physics 111: Lecture 25, Pg 7

θ

Initial Conditions... So we find φ = -π/2!! x(t) = A cos(ωt - π/2 ) v(t) = -ωA sin(ωt - π/2 ) a(t) = -ω2A cos(ωt - π/2 )

x(t) = A sin(ωt) v(t) = ωA cos(ωt) a(t) = -ω2A sin(ωt)

A

x(t) π

k

m

0

2π

-A x Physics 111: Lecture 25, Pg 8

ωt

Lecture 25, Act 2 Initial Conditions ●

A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describes its velocity and acceleration as a function of time?

(a) v(t) = -vmax sin(ωt)

a(t) = -amax cos(ωt)

(b) v(t) = vmax sin(ωt)

a(t) = amax cos(ωt)

(c) v(t) = vmax cos(ωt)

k t=0

a(t) = -amax cos(ωt)

y m

d 0

(both vmax and amax are positive numbers) Physics 111: Lecture 25, Pg 9

Lecture 25, Act 2 Solution Since we start with the maximum possible displacement at t = 0 we know that: y = d cos(ωt) v=

dy = −ωd sin ( ωt ) ≡ −v max sin ( ωt ) dt

dv a= = −ω2d cos( ωt ) ≡ −amax cos ( ωt ) dt

k t=0

y m

d 0

Physics 111: Lecture 25, Pg 10

Review of Simple Pendulum ●

Using τ = Iα and sin θ ≈ θ for small θ

z

d 2θ − mgLθ = mL dt 2 2

τ

I

α

θ

We found d 2θ 2 = − ω θ 2 dt

where

ω=

L

g L

m

Which has SHM solution θ = θ0 cos(ωt + φ)

d mg

Physics 111: Lecture 25, Pg 11

Review of Rod Pendulum ●

Using τ = Iα and sinθ ≈ θ for small θ

z

2 L 1 2 d θ − mg θ = mL 2 3 dt 2

τ

I

L/2 θ

α

xCM

We found d 2θ 2 = − ω θ 2 dt

where

ω=

L d mg

3g 2L

Which has SHM solution θ = θ0 cos(ωt + φ)

Physics 111: Lecture 25, Pg 12

General Physical Pendulum ●

●

Suppose we have some arbitrarily shaped solid of mass M hung on a fixed axis, and that we know where the CM is located and what the moment of inertia I about the axis is. The torque about the rotation (z) axis for small θ is (sin θ ≈ θ ) d 2θ − MgRθ = I τ = ≈ dt 2 -Mgd -MgRθ τ

2

d θ dt 2

= −ω 2 θ

where

ω=

α

MgR I

Physical Pendulum z-axis R θ xCM d Mg

θ = θ0 cos(ωt + φ) Physics 111: Lecture 25, Pg 13

Lecture 25, Act 3 Physical Pendulum ●

A pendulum is made by hanging a thin hoola-hoop of diameter D on a small nail. ➧ What is the angular frequency of oscillation of the hoop for small displacements? (ICM = mR2 for a hoop)

(a)

ω=

g D

(b)

ω=

2g D

(c)

ω=

g 2D

pivot (nail)

D

Physics 111: Lecture 25, Pg 14

Lecture 25, Act 3 Solution ●

Hoop Pendulum

The angular frequency of oscillation of the hoop for small displacements will be given by ω =

mgR (see Lecture 25 notes) I

Use parallel axis theorem: I = Icm + mR2 = mR2 + mR2 = 2mR2 pivot (nail) ω=

mgR g g = = 2R D 2 mR 2

So

ω=

g D

cm x R

m

Physics 111: Lecture 25, Pg 15

Torsion Pendulum ●

●

Consider an object suspended by a wire attached at its CM. The wire defines the rotation axis, and the moment of inertia I about this axis is known. The wire acts like a “rotational spring.” ➧ When the object is rotated, the wire is twisted. This produces a torque that opposes the rotation. ➧ In analogy with a spring, the torque produced is proportional to the displacement: τ = -kθ

wire θ

τ I

Physics 111: Lecture 25, Pg 16

Torsion Pendulum... ●

Torsion Pendulum

Since τ = -kθ, τ = Iα becomes −kθ = I

2

d θ dt

2

2

wire

d 2θ dt 2

= −ω θ

where

ω=

θ

τ

k I

I

This is similar to the “mass on spring” except I has taken the place of m (no surprise).

Physics 111: Lecture 25, Pg 17

Energy in SHM ●

For both the spring and the pendulum, we can derive the SHM solution by using energy conservation.

●

The total energy (K + U) of a system undergoing SHM will always be constant!

●

This is not surprising since there are only conservative forces present, hence K+U energy is conserved.

U E -A

0

K U A

Physics 111: Lecture 25, Pg 18

s

SHM and quadratic potentials ● ● ●

SHM will occur whenever the potential is quadratic. Generally, this will not be the case: For example, the potential between H atoms in an H2 molecule looks U something like this: E

U -A

0

K U A

x

Physics 111: Lecture 25, Pg 19

x

SHM and quadratic potentials... However, if we do a Taylor expansion of this function about the minimum, we find that for small displacements, the potential IS quadratic: U U(x) = U(x0 ) + U′(x0 ) (x- x0 ) 1 + U′′ (x0 ) (x- x0 )2+.... 2

U x0

U′(x0) = 0 (since x0 is minimum of potential) x′ = x - x0 and U(x0 ) = 0 1 Then U(x) = 2 U′′ (x0 ) x′ 2 Define

x x′

Physics 111: Lecture 25, Pg 20

SHM and quadratic potentials...

1 U(x) = U′′ (x0) x′ 2 2 Let k = U′′ (x0)

U U x0

Then: 1 U(x) = 2 k x′ 2

x x′

SHM potential!! Physics 111: Lecture 25, Pg 21

Problem: Vertical Spring ●

A mass m = 102 g is hung from a vertical spring. The equilibrium position is at y = 0. The mass is then pulled down a distance d = 10 cm from equilibrium and released at t = 0. The measured period of oscillation is T = 0.8 s. ➧ What is the spring constant k? ➧ Write down the equations for the position, velocity, and acceleration of the mass as functions of time. ➧ What is the maximum velocity? ➧ What is the maximum acceleration?

k y 0 m t=0

-d

Physics 111: Lecture 25, Pg 22

Problem: Vertical Spring...

●

What is k ?

So: k = 7 .85 s

k = ω2m

2π = 7 .85 s −1 T

ω=

(

k m

ω=

)

−1 2

k y

N 0 .102 kg = 6 .29 m

0 m t=0

-d

Physics 111: Lecture 25, Pg 23

Problem: Vertical Spring... ●

What are the equations of motion?

●

At t = 0, ➧ y = -d = -ymax

k

➧ v=0 ●

So we conclude:

y y(t) = -d cos(ωt) v(t) = ωd sin(ωt) a(t) = ω2d cos(ωt)

0 m t=0

-d

Physics 111: Lecture 25, Pg 24

Problem: Vertical Spring... y(t) = -d cos(ωt) v(t) = ωd sin(ωt) a(t) = ω2d cos(ωt)

0

π

2π

ωt k y

xmax = d = .1m

0

vmax = ωd = (7.85 s-1)(.1m) = 0.78 m/s amax = ω d = (7.85 s ) (.1m) = 6.2 m/s 2

-1 2

2

m t=0

-d

Physics 111: Lecture 25, Pg 25

Transport Tunnel ●

A straight tunnel is dug from Urbana through the center of the Earth and out the other side. A physics 111 student jumps into the hole at noon. ➧ What time does she get back to Urbana?

Physics 111: Lecture 25, Pg 26

Transport Tunnel... FG ( R ) =

GmM R R2

where MR is the FG

mass inside radius R

R RE

MR

FG ( R ) M R RE2 = FG ( RE ) R 2 M E

but

MR ∝ R 3

FG ( R ) R 3 RE2 R = 2 3 = FG ( RE ) R RE RE

Physics 111: Lecture 25, Pg 27

Transport Tunnel... FG ( R )

FG ( RE )

FG

=

R RE

FG ( RE ) = −mg

R RE

MR

R FG = −mg = −kR RE

Like a mass on a spring with

k =

mg RE

Physics 111: Lecture 25, Pg 28

Transport Tunnel... Like a mass on k = mg RE a spring with FG

So: ω =

R RE

MR

k g = m RE

plug in g = 9.81 m/s2 and RE = 6.38 x 106 m get ω = .00124 s-1 2π and so T = ω = 5067 s ≈ 84 min Physics 111: Lecture 25, Pg 29

Transport Tunnel... ●

So she gets back to Urbana 84 minutes later, at 1:24 p.m.

Physics 111: Lecture 25, Pg 30

Transport Tunnel... ●

Strange but true: The period of oscillation does not require that the tunnel be straight through the middle!! Any straight tunnel gives the same answer, as long as it is frictionless and the density of the Earth is constant.

Physics 111: Lecture 25, Pg 31

Transport Tunnel... ●

Another strange but true fact: An object orbiting the earth near the surface will have a period of the same length as that of the transport tunnel. a = ω 2R 9.81 = ω2 6.38(10)6 m ω = .00124 s-1 so T =

2π = 5067 s ω ≈ 84 min Physics 111: Lecture 25, Pg 32

Simple Harmonic Motion: Summary ω=

Force:

d 2s 2 = −ω s 2 dt

k k m

s 0 m

k

m s

0 Solution: s = A cos(ωt + φ)

ω=

g L

s

L

Physics 111: Lecture 25, Pg 33

Recap of today’s lecture ● ● ● ● ●

● ● ● ●

Recap of last lecture Using “initial conditions” to solve problems (Text: 14-1) The general physical pendulum (Text: 14-3) The torsion pendulum Energy in SHM (Text: 14-2) ➧ Atomic Vibrations Problem: Vertical Spring (Text: 14-3) Problem: Transport Tunnel SHM Review Look at textbook problems Chapter 14: # 51, 53, 57, 58, 65, 125 Physics 111: Lecture 25, Pg 34