PHYS 1443 – Section 004 Lecture #25

Tuesday, Dec. 2, 2014 PHYS 1443-004, Fall 2014 Dr. Jaehoon Yu

1

PHYS 1443 – Section 004 Lecture #25

Tuesday, Dec. 2, 2014 Dr. Jaehoon Yu

• SHM & Uniform Circular Motion • Pendulum

•  Simple Pendulum •  Physical Pendulum •  Torsion Pendulum

• Damped Oscillation & Resonance • Superposition and Interference • Waves on Strings

Tuesday, Dec. 2, 2014 2

Announcements •  Final comprehensive exam

–  9:00 – 10:30am, next Thursday, Dec. 11 in this room –  Covers CH 1.1 through what we finish Tuesday, Dec. 2 plus the

math refresher –  Mixture of multiple choice and free response problems –  Bring your calculator but DO NOT input formula into it!

•  Your phones or portable computers are NOT allowed as a replacement! –  You can prepare a one 8.5x11.5 sheet (front and back) of

handwritten formulae and values of constants for the exam •  None of the parts of the solutions of any problems •  No derived formulae, derivations of equations or word definitions!

–  Do NOT Miss the exam!

PHYS 1443-004, Fall 2014 Dr. Jaehoon Yu


3

Simple Harmonic and Uniform Circular Motions Uniform circular motion can be understood as a superposition of two simple harmonic motions in x and y axis.

When the particle rotates at a uniform angular speed ϖ, x and y coordinate position become

Since the linear velocity in a uniform circular motion is Aϖ, the velocity components are

xt=0

x

y

O

P φ A

x

y

O

P

θ A

Q

ϖ

x y

t=t θ=ϖt+φ

x

y

O

P

θ A

Q

v

vx

xv

x

y

O

P

θ A

Q

a

ax

Since the radial acceleration in a uniform circular motion is v2/A=ϖ2A, the components are

xa

θcosA= = Acos ωt +ϕ( )y θsinA= = Asin ωt +ϕ( )

θsinv−= = −Aω sin ωt +ϕ( )yv θcosv+= = Aω cos ωt +ϕ( )

θcosa−= = −Aω 2 cos ωt +ϕ( )ya θsina−= = −Aω 2 sin ωt +ϕ( )


4

The Period and Sinusoidal Nature of SHM

v0

Consider an object moving on a circle with a constant angular speed ϖ

If you look at it from the side, it looks as though it is doing a SHM

sinθ2

0 1xv vA

⎛ ⎞= −⎜ ⎟⎝ ⎠

0v0

2 ATvπ=

20

12mv

2 mTk

π= f

Since it takes T to complete one full circular motion From an energy relationship in a spring SHM

0kv Am

=

Thus, T is

Natural Frequency

0

vv

=2 2A xA−=

2

1 xA

⎛ ⎞= −⎜ ⎟⎝ ⎠2 ATπ= 2 Afπ=

212kA=

1T

=12

kmπ

=


5

Example for Uniform Circular Motion A particle rotates counterclockwise in a circle of radius 3.00m with a constant angular speed of 8.00 rad/s. At t=0, the particle has an x coordinate of 2.00m and is moving to the right. A) Determine the x coordinate as a function of time.

Since the radius is 3.00m, the amplitude of oscillation in x direction is 3.00m. And the angular frequency is 8.00rad/s. Therefore the equation of motion in x direction is

Since x=2.00, when t=0

However, since the particle was moving to the right φ=- 48.2o,

Using the displcement

x

2.00 = 3.00m( )cosϕ;

x = 3.00m( )cos 8.00t − 48.2( )

Find the x components of the particle’s velocity and acceleration at any time t.

vx

Likewise, from velocity ax

= Acosθ= 3.00m( )cos 8.00t +ϕ( )

ϕ = cos−1 2.00

3.00⎛⎝⎜

⎞⎠⎟ = 48.2

= dxdt

= − 3.00 ⋅8.00( )sin 8.00t − 48.2( ) = −24.0m / s( )sin 8.00t − 48.2( )

= dvdt

= −24.0 ⋅8.00( )cos 8.00t − 48.2( ) = −192m / s2( )cos 8.00t − 48.2( )


6

The Pendulum A simple pendulum also performs a periodic motion.

The net force exerted on the bob is ∑ rF

s Lθ=

Again became a second order differential equation, satisfying the condition for a simple harmonic motion

If θ is very small, sinθ~θ

Since the arc length, s, is

2

2

dtsd d 2θ

dt 2results mg

m

θ

L

T

s

d 2θdt 2 L

g=ωgiving angular frequency

The period for this motion is T The period only depends on the length of the string and the gravitational acceleration

= T −mgcosθ 0=

Ft∑ = −mgsinθ ma= = m d 2sdt 2

= L d2θdt 2

θsing−= = − gLsinθ

θLg−= θω 2−=

ωπ2=

gLπ2=


7

Example for A Simple Pendulum Grandfather clock. (a) Estimate the length of the pendulum in a grandfather clock that ticks once per second.

Since the period of a simple pendulum motion is T

The length of the pendulum in terms of T is

L = T2g4π 2

Thus the length of the pendulum when T=1s is

L = T 2g4π 2

gLπ2=

(b) What would be the period of the clock with a 1m long pendulum?

gLπ2=T 1.02 2.0

9.8sπ= =

2

1 9.8 0.254

mπ×= =


8

Example for Pendulum Christian Huygens (1629-1695), the greatest clock maker in history, invented the clock using the simple pendulum in 1656. He suggested that an international unit of length could be defined as the length of a simple pendulum having a period of exactly 1s. How much shorter would our length unit be had this suggestion been followed?

Since the period of a simple pendulum motion is T

The length of the pendulum in terms of T is 2

2

4πgTL =

Thus the length of the pendulum when T=1s is

L = T2g4π 2 =

Therefore the difference in length with respect to the current definition of 1m is LΔ

ωπ2=

gLπ2=

L−=1 m752.0248.01 =−=

1× 9.84π 2 = 0.248m


9

Physical Pendulum Physical pendulum is an object that oscillates about a fixed axis which does not go through the object’s center of mass.

Therefore, one can rewrite

Thus, the angular frequency ω is

The magnitude of the net torque provided by the gravity is ∑τ

ω = mgdI

And the period for this motion is T =

By measuring the period of physical pendulum, one can measure moment of inertia.

O

CM d

dsinθ θ

mg

Consider a rigid body pivoted at a point O that is a distance d from the CM.

∑τThen

d 2θdt 2

Does this work for simple pendulum?

= −mgd sinθ

αI= = I d2θdt 2

= −mgd sinθ

= − mgdIsinθ ≈ − mgd

I⎛⎝⎜

⎞⎠⎟θ = −ω 2θ

2πω

=2πI

mgd


10

Example for Physical Pendulum A uniform rod of mass M and length L is pivoted about one end and oscillates in a vertical plane. Find the period of oscillation if the amplitude of the motion is small.

Moment of inertia of a uniform rod, rotating about the axis at one end is

Since L=1m, the period is

L

O Pivot

CM

Mg

2

31MLI =

The distance d from the pivot to the CM is L/2, therefore the period of this physical pendulum is

T

Calculate the period of a meter stick that is pivot about one end and is oscillating in a vertical plane.

T = So the frequency is

161.01 −== sT

f

ωπ2=

MgdIπ2=

MgLML322

2

π=gL322π=

2π 2L3g

=2π 23⋅9.8

= 1.64s


11

Torsion Pendulum When a rigid body is suspended by a wire to a fixed support at the top and the body is twisted through some small angle θ, the twisted wire can exert a restoring torque on the body that is proportional to the angular displacement.

Applying the Newton’s second law of rotational motion

Thus, the angular frequency ω is

The torque acting on the body due to the wire is

τ

Iκω =

And the period for this motion is T

This result works as long as the elastic limit of the wire is not exceeded

∑τ

κ is the torsion constant of the wire

2

2

dtd θ

O

P θmax

Then, again the equation becomes

κθ−=αI=

2

2

dtdI θ= κθ−=

= − κI

⎛⎝⎜

⎞⎠⎟θ = −ω 2θ

ωπ2=

κπ I2=


12

Damped Oscillation More realistic oscillation where an oscillating object loses its mechanical energy in time by a retarding force such as friction or air resistance.

How do you think the motion would look?

Amplitude gets smaller as time goes on since its energy is spent.

Types of damping A: Underdamped B: Critically damped C: Overdamped


13

Forced Oscillation; Resonance When a vibrating system is set into motion, it oscillates with its natural frequency f0.

However a system may have an external force applied to it that has its own particular frequency (f), causing a forced vibration.

For a forced vibration, the amplitude of vibration is found to be dependent on the difference between f and f0. and is maximum when f=f0.

A: light damping B: Heavy damping The amplitude can be large when f=f0, as long as damping is small.

This is called the resonance. The natural frequency f0 is also called resonant frequency.

f0 =

12π

km


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Superposition and Interference Superposition Principle

If two or more traveling waves are moving through a medium, the resultant wave function at any point is the algebraic sum of the wave functions of the individual waves. The waves that follow this principle are called linear waves which in general have small amplitudes. The ones that don’t are nonlinear waves with larger amplitudes.

What do you think will happen to the water waves when you throw two stones on the pond?

Two traveling linear waves can pass through each other without being destroyed or altered.

What happens to the waves at the point where they meet?

They will pass right through each other.

The shape of wave will changeè Interference

Thus, one can write the resultant wave function as y = y1 + y2 + ⋅⋅⋅+ yn = yi

i=1

n

∑

Constructive interference: The amplitude increases when the waves meet Destructive interference: The amplitude decreases when the waves meet


15

Speed of Waves on Strings How do we determine the speed of a transverse pulse traveling on a string? If a string under tension is pulled sideways and released, the tension is responsible for accelerating a particular segment of the string back to the equilibrium position.

The speed of the wave increases.

So what happens when the tension increases?

Which law does this hypothesis based on?

Based on the hypothesis we have laid out above, we can construct a hypothetical formula for the speed of wave

For the given tension, acceleration decreases, so the wave speed decreases. Newton’s second law of motion

The acceleration of the particular segment increases

v = Tµ

Which means?

Now what happens when the mass per unit length of the string increases?

T: Tension on the string m: Unit mass per length

Is the above expression dimensionally sound? T=[MLT-2], m=[ML-1] (T/m)1/2=[L2T-2]1/2=[LT-1]


16

Speed of Waves on Strings cont’d Let’s consider a pulse moving to right and look at it in the frame that moves along with the the pulse.

Since in the reference frame moves with the pulse, the segment is moving to the left with the speed v, and the centripetal acceleration of the segment is

What is the mass of the segment when the line density of the string is µ?

Using the radial force component

Now what do the force components look in this motion when θ is small?

T T Fr

O

θ θ

θ θ

Ds v

R ar =v2

R

Ft∑ = T cosθ −T cosθ =

m =

Fr∑ =

Therefore the speed of the pulse is µTv =

Fr∑ = 2T sinθ ≈ 2Τθ

0

µΔs = µR2θ =2µRθ

ma =mv2

R= 2µRθ v

2

R= 2Tθ


17

Example A uniform cord has a mass of 0.300kg and a length of 6.00m. The cord passes over a pulley and supports a 2.00kg object. Find the speed of a pulse traveling along this cord.

Thus the speed of the wave is

Since the speed of wave on a string with line density m and under the tension T is

T = Mg = 2.00 × 9.80 = 19.6kg ⋅m / s2

M=2.00kg

1.00m 5.00m v = T

µ

The line density µ is µ = 0.300kg6.00m

= 5.00 ×10−2 kg /m

The tension on the string is provided by the weight of the object. Therefore

v = Tµ= 19.6

5.00 ×10−2 = 19.8m / s


18

Reflection and Transmission A pulse or a wave undergoes various changes when the medium it travels changes. Depending on how rigid the support is, two radically different reflection patterns can be observed. 1.  The support is rigidly fixed: The reflected pulse will be inverted to the

original due to the force exerted on to the string by the support in reaction to the force on the support due to the pulse on the string.

2.  The support is freely moving: The reflected pulse will maintain the original shape but moving in the reverse direction.

If the boundary is intermediate between the above two extremes, part of the pulse reflects, and the other undergoes transmission, passing through the boundary and propagating in the new medium. When a wave pulse travels from medium A to B: •  vA> vB (or mA<mB), the pulse is inverted upon reflection. •  vA< vB(or mA>mB), the pulse is not inverted upon reflection.


19

Sinusoidal Waves Equation of motion of a simple harmonic oscillation is a sine function.

But it does not travel. Now how does wave form look like when the wave travels?

Thus the wave form can be rewritten

By definition, the speed of wave in terms of wave length and period T is

y = Asin 2πλx⎛

⎝⎜⎞⎠⎟

Defining, angular wave number k and angular frequency w,

k ≡ 2πλ; ω = 2π

T

The function describing the position of particles, located at x, of the medium through which the sinusoidal wave is traveling can be written at t=0

v = λT

Wave Length

The wave form of the wave traveling at the speed v in +x at any given time t becomes y = Asin 2π

λx − vt( )⎛

⎝⎜⎞⎠⎟

Amplitude

y = Asin 2π xλ− tT

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

y = Asin kx −ωt( )The wave form becomes

Frequency, f, f = 1T

Wave speed, v

v = λT= ωk

General wave form

y = Asin kx −ωt +ϕ( )


20

Example A sinusoidal wave traveling in the positive x direction has an amplitude of 15.0cm, a wavelength of 40.0cm, and a frequency of 8.00Hz. The vertical displacement of the medium at t=0 and x=0 is also 15.0cm. a) Find the angular wave number k, period T, angular speed ϖ, and speed v of the wave.

At x=0 and t=0, y=15.0cm, therefore the phase φ becomes

Using the definition, angular wave number k is

b) Determine the phase constant φ, and write a general expression of the wave function.

mradk /7.1500.540.022 ==== ππ

λπ

sradfT

/3.5022 === ππωPeriod is sec125.000.811 ===

fT Angular

frequency is

Using period and wave length, the wave speed is smfT

v /2.300.8400.0 =×=== λλ

y = 0.150sin ϕ( ) = 0.150

sinϕ = 1; ϕ = π2

Thus the general wave function is y = Asin kx −ωt +ϕ( ) = 0.150sin 15.7x − 50.3t + π

2⎛⎝⎜

⎞⎠⎟


21

Sinusoidal Waves on a String Let’s consider the case where a string is attached to an arm undergoing a simple harmonic oscillation. The trains of waves generated by the motion will travel through the string, causing the particles in the string to undergo simple harmonic motion on y-axis.

This wave function describes the vertical motion of any point on the string at any time t. Therefore, we can use this function to obtain transverse speed, vy, and acceleration, ay.

y = Asin 2πλx⎛

⎝⎜⎞⎠⎟

The maximum speed and the acceleration of the particle in the medium at position x at time t are

If the wave at t=0 is What does this mean?

y = Asin kx −ωt( )

ϕ = 0

The wave function can be written

vy =dydt xconst

= ∂y∂t

= −ωAcos kx −ωt( ) ay =dvydt xconst

=∂vy∂t

= −ω 2Asin kx −ωt( )

vy,max =ωA

ay,max =ω2A

How do these look for simple harmonic motion?

These are the speed and acceleration of the particle in the medium not of the wave.


22

Example A string is driven at a frequency of 5.00Hz. The amplitude of the motion is 12.0cm, and the wave speed is 20.0m/s. Determine the angular frequency w and angular wave number k for this wave, and write and expression for the wave function.

Using frequency, the angular frequency is

Thus the general expression of the wave function is

Angular wave number k is

ω = 2πΤ

= 2π f = 2π ⋅5.00 = 31.4rad / s

k = 2πλ

= 2πvT

= 2π fv

= ωv= 31.420.0

= 1.57rad /m

y = Asin kx −ωt( ) = 0.120sin 1.57x − 31.4t( )


23

Rate of Energy Transfer by Sinusoidal Wave on a String Waves traveling through medium carries energy.

When an external source performs work on the string, the energy enters into the string and propagates through the medium as wave.

What is the potential energy of one wave length of a traveling wave? Δx, Δm Elastic potential energy of a particle in a simple harmonic motion U = 1

2ky2

Since w2=k/m U = 12mω 2y2 The energy ΔU of the segment Δm is ΔU = 1

2Δmω 2y2 = 1

2µΔxω 2y2

As Δxà0, the energy ΔU becomes dU = 12µω 2y2dx

Using the wave function,the energy is dU = 12µω 2A2 sin2 kx −ωt( )dx

For the wave at t=0, the potential energy in one wave length, l, is

Uλ =12µω 2A2 sin2 kxdx

x=0

x=λ

∫ = 12µω 2A2 1− cos2kx

2dx

x=0

x=λ

∫

Recall k=2π/l = 12µω 2A2 1

2x − 1

4ksin 4π x

λ⎡⎣⎢

⎤⎦⎥x=0

x=λ

= 14µω 2A2λ


24

Rate of Energy Transfer by Sinusoidal Waves cont’d How does the kinetic energy of each segment of the string in the wave look?

Since the vertical speed of the particle is

As Δxà0, the energy ΔK becomes dK = 12µω 2A2 cos2 kx −ωt( )dx

For the wave at t=0, the kinetic energy in one wave length, l, is

Kλ =12µω 2A2 cos2 kxdx

x=0

x=λ

∫ = 12µω 2A2 1+ cos2kx

2dx

x=0

x=λ

∫

Recall k=2π/l

vy = −ωAcos kx −ωt( )The kinetic energy, ΔK, of the segment Δm is

ΔK = 12Δmvy

2 = 12µΔxω 2A2 cos2 kx −ωt( )

Just like harmonic oscillation, the total mechanical energy in one wave length, l, is Eλ =Uλ + Kλ =

12µω 2A2λ

As the wave moves along the string, the amount of energy passes by a given point changes during one period. So the power, the rate of energy transfer becomes

P = Eλ

Δt= 12µω 2A2 λ

T P of any sinusoidal wave is proportion to the square of angular frequency, the square of amplitude, density of medium, and wave speed.

= 12µω 2A2 1

2x + 1

4ksin 4π x

λ⎡⎣⎢

⎤⎦⎥x=0

x=λ

= 14µω 2A2λ

= 12µω 2A2v


25

Example A taut string for which µ=5.00x10-2 kg/m is under a tension of 80.0N. How much power must be supplied to the string to generate sinusoidal waves at a frequency of 60.0Hz and an amplitude of 6.00cm?

The speed of the wave is

Since the rate of energy transfer is

v = Tµ= 80.0

5.00 ×10−2 = 40.0m / s

Using the frequency, angular frequency ϖ is

ω = 2πΤ

= 2π f = 2π ⋅60.0 = 377rad / s

P = Eλ

Δt= 12µω 2A2v

= 12× 5.00 ×10−2 × 377( )2 × 0.06( )2 × 40.0( ) = 512 W( )

Congratulations!!!!

I certainly had a lot of fun with ya’ll and am truly proud of you!

You all are impressive and have done very well!!!

Good luck with your exam!!!

Have great holidays!! Tuesday, Dec. 2, 2014 PHYS 1443-004, Fall 2014

Dr. Jaehoon Yu 26

PHYS 1443 – Section 004 Lecture #25

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