Simple Harmonic Oscillator

8.01

Week 13D1

Today’s Reading Assignment Young and Freedman: 14.1-14.6

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Announcements

Math Review Tuesday Nov 29 from 9-11 pm

Pset 11 due Thursday Dec 1 at 9 pm

W013D2 Reading Assignment Young and Freedman: 14.1-14.6

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Mass on a Spring C2:Simple Harmonic Motion

Hooke’s Law

Define system, choose coordinate system.

Draw free-body diagram.

Hooke’s Law

springˆkxF i

2

2

d xkx m

dt

5

Mass on a Spring

F =−kx(1) (2)

(3) (4)

Concept Question: Simple Harmonic Oscillator

Which of the following functions x(t) has a second derivative which is proportional to the negative of the function

x(t) =Acos

2πTt

⎛

⎝⎜⎞

⎠⎟

x(t) =Ae−t/T

x(t) =Aet/T x(t) =

12at2

2

2?

d xx

dt∝ −

1.

2.

3.

4.

Concept Question Answer: Simple Harmonic Oscillator

Answer 4. By direct calculation, when

x(t) =Acos

2πTt

⎛

⎝⎜⎞

⎠⎟

dx(t)

dt=−

2πT

⎛

⎝⎜⎞

⎠⎟Asin

2πTt

⎛

⎝⎜⎞

⎠⎟

d 2x(t)

dt2=−

2πT

⎛

⎝⎜⎞

⎠⎟

2

Acos2πTt

⎛

⎝⎜⎞

⎠⎟=−

2πT

⎛

⎝⎜⎞

⎠⎟

2

x(t)

Demo slide: spray paint oscillator C4

Illustrating choice of alternative representations for position as a function of time (amplitude and phase or sum of sin and cos)

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Simple Harmonic Motion

x(t) =Ccos(ω0t) + Dsin(ω0t) −kx=m

d2xdt2

⇒

ω0 = k / m=2π f =2π / T

x(t) =Acos(ω0t+φ) ⇒

C =Acos(φ)D=−Asin(φ)

A = C2 + D2

tanφ =−D / C

Simple Harmonic Motion

Equation of Motion:

Solution: Oscillatory with Period

Position:

Velocity:

Initial Position at t = 0:

Initial Velocity at t = 0:

General Solution:

2

2

d xkx m

dt

x =Ccos(ω0t) + Dsin(ω0t)

v

x=dxdt

=−ω0Csin(ω0t) +ω0Dcos(ω0t)

x0≡x(t=0) =C

v

x ,0≡vx(t=0) =ω0D

x =x0 cos(ω0t) +ω0vx,0 sin(ω0t)

T =2π / ω0 =2π m/ k

Worked Example: Simple Harmonic Motion Block-Spring

A block of mass m , attached to a spring with spring constant k, is free to slide along a horizontal frictionless surface. At t = 0 the block-spring system is stretched an amount x0 > 0 from the equilibrium position and released from rest. What is the x -component of the velocity of the block when it first comes back to the equilibrium?

Worked Example: Simple Harmonic Motion Block-Spring

At t = 0, the block is at x(t = 0) = x0, and it is released from rest so vx(t = 0) = 0. Therefore the initial conditions are

So the position and x-component of the velocity of the block are given by of the block as a function of time is

The particle first returns to equilibrium at t = T/4, therefore

sin(ω0T / 4) =1 ⇒ vx =−ω0x0 =− k / m( )x0

x(t =0) =C =x0 vx(t =0) =ω0D=0

x(t) =x0 cos(ω0t) vx(t) =−ω0x0 sin(ω0t)

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Mass on a Spring: Energy

x(t) =Acos(ω0t+φ)

Constant energy oscillates between kinetic and potential energies

K(t) =(1 / 2)m(vx(t))2 =(1 / 2)kA2 sin2 (ω0t+φ)

U (t) =(1 / 2)kx2 =(1 / 2)kA2cos2(ω0t+φ)

vx(t) =−ω0Asin(ω0t+φ)

ω0 = k / m

A = x0

2 + vx2 / ω0

2

E =K (t) +U (t) =(1 / 2)kA2 =(1 / 2)kx02 + (1 / 2)mvx

2 =constant

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Energy and Simple Harmonic Motion

dE / dt =0 ⇒

0 =kxdxdt

+mvxdvxdt

0 =kx+mx

d2xdt2

Concept Question: Simple Harmonic Motion Velocity

1. 2.

3. 4.

A block of mass m is attached to a spring with spring constant k is free to slide along a horizontal frictionless surface.At t = 0 the block-spring system is compressed an amount x0 < 0 from the equilibrium position and is released from rest. What is the x -component of the velocity of the block when it first comes back to the equilibrium?

v

x=−x0

T4

vx=x0

T4

v

x=−

kmx0

v

x=

kmx0

CQ Answer: Simple Harmonic Motion Velocity

Answer 3. The particle starts with potential energy. When it first returns to equilibrium it now has only kinetic energy. Since the energy of the block-spring system is constant:

Take positive root because object is moving in positive x-direction when it first comes back to equilibrium, and xo < 0, we require that vx,o > 0, therefore

v

x=−

kmx0

(1 / 2)mvx2 =(1 / 2)kx0

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Table Problem: Block-Spring System with No Friction

A block of mass m slides along a frictionless horizontal surface with speed v x,0. At t = 0 it hits a spring with spring constant k and begins to slow down. How far is the spring compressed when the block has first come momentarily to rest?

Kinetic Energy vs. Potential Energy

StateKinetic energy

Potential energy

Mechanical energy

Initial

Final

U

0=

12kx0

2 E0

=K0 +U0

K

f=

12mvx, f

2 E

f=K f +U f =E0

v

x ,0≠0

v

x , f≠0

x

f≠0

x0≠0

K

0

12mvx,0

2

U

f

1

2kx

f2

Source

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Energy DiagramChoose zero point for potential energy:

Potential energy function:

Mechanical energy is represented by a horizontal line

U (x)

1

2kx2 , U (x 0) 0

U (x 0) 0

Emechanical K(x) U(x) 1

2mvx

2 1

2kx2

K Emechanical UGraph of Potential energy function

U(x) vs. x

Concept Question: Harmonic Energy Diagram

The position of a particle is given by

Where was the particle at t = 0?

1) 12) 23) 34) 45) 56) 1 or 57) 2 or 4

( )( ) cos( ) sin , 0x t D t D t Dω ω= − >

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Concept Question: Harmonic Energy Diagram Answer

The position and x-component of velocity of a particle is given by

Hence

Thus the particle starts out with a positive x- coordinate and the initial x-component of the velocity is negative therefore it is moving toward the origin. It has both non-zero potential and kinetic energies and hence it is not at it’s maximal position so at it is located at position 4.

( )( ) cos( ) sin , 0x t D t D t Dω ω= − >

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v

x ,0=−ω0D < 0

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Oscillating Systems:Simple Harmonic Motion

Demo: Air Spring with Oscillating steel ball C12

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Worked Example: fluid oscillations in a U-tube

A U-tube open at both ends to atmospheric pressure is filled with an incompressible fluid of density . The cross-sectional area A of the tube is uniform and the total length of the column of fluid is L. A piston is used to depress the height of the liquid column on one side by a distance x, and then is quickly removed. What is the frequency of the ensuing simple harmonic motion? Assume streamline flow and no drag at the walls of the U-tube. The gravitational constant is g.

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Demonstration: U-tube oscillations

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Table Problem: Earthquake

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In 1993, to try to protect its historic courthouse from damage due to earthquakes, the city of San Francisco designed a special foundation. The building rests on N identical supports. Each support has a small diameter spherical cap anchored to the ground resting in an inverted spherical cup of radius R attached to the building. As a result, if the building undergoes a small horizontal displacement x, it rises by an amount as shown in the diagram. Building has mass M.

a) Assume that there is no friction. Find the differential equation which describes the dynamics of small displacements x(t) of the center of the cups from the centers of the caps.

b) A sudden shock sets the building moving with the initial conditions x(0) = 0 and . Find an expression for x(t) for all subsequent times.

Δh ≈ x2 / 2R

vx (0) =v0