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Physics 151: Lecture 35 Today’s Agenda. Topics Waves on a string Superposition Power.  and T are related !. Travleing 1-D wave: y(x,t):. Review: Wave Properties.

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physics 151 lecture 35 today s agenda
Physics 151: Lecture 35 Today’s Agenda
  • Topics
    • Waves on a string
    • Superposition
    • Power
slide2

 and T are related !

  • Travleing 1-D wave: y(x,t):

Review: Wave Properties...

  • The speed of a wave (v) is a constant and depends only on the medium, not on amplitude (A), wavelength () or period (T).
  • remember : T = 1/ f and T = 2 / 
slide3

Example

  • Bats can detect small objects such as insects that are of a size on the order of a wavelength. If bats emit a chirp at a frequency of 60 kHz and the speed of soundwaves in air is 330 m/s, what is the smallest size insect they can detect ?
    • a. 1.5 cm
    • b. 5.5 cm
    • c. 1.5 mm
    • d. 5.5 mm
    • e. 1.5 um
    • f. 5.5 um
slide4

Example

  • Write the equation of a wave, traveling along the +x axis with an amplitude of 0.02 m, a frequency of 440 Hz, and a speed of 330 m/sec.
    • A. y = 0.02 sin [880 (x/330 – t)]
    • b.y = 0.02 cos [880 x/330 – 440t]
    • c.y = 0.02 sin [880(x/330 + t)]
    • d.y = 0.02 sin [2(x/330 + 440t)]
    • e.y = 0.02 cos [2(x/330 - 440t)]
slide5

Example

  • For the transverse wave described by

y = 0.15 sin [ p(2x - 64 t)/16] (in SI units),

determine the maximum transverse speed of the particles of the medium.

    • a. 0.192 m/s
    • b. 0.6 m/s
    • c. 9.6 m/s
    • d. 4 m/s
    • e. 2 m/s
lecture 34 act 4 wave motion
Lecture 34, Act 4Wave Motion
  • A heavy rope hangs from the ceiling, and a small amplitude transverse wave is started by jiggling the rope at the bottom.
  • As the wave travels up the rope, its speed will:

v

(a) increase

(b) decrease

(c) stay the same

  • Can you calcuate how long will it take for a pulse travels a rope of length L and mass m ?
superposition

See text: 16.4

Superposition
  • Q: What happens when two waves “collide” ?
  • A: They ADD together!
    • We say the waves are “superposed”.

Animation-1

Animation-2

see Figure 16.8

aside why superposition works
Aside: Why superposition works
  • It can be shown that the equation governing waves (a.k.a. “the wave equation”) is linear.
    • It has no terms where variables are squared.
  • For linear equations, if we have two (or more) separate solutions, f1 and f2 , then Bf1+Cf2 is also a solution !
  • You have already seen this in the case of simple harmonic motion:

linear in x !

x = Bsin(t)+Ccos(t)

superposition interference

will add constructively

will add destructively

See text: 16.4

Superposition & Interference
  • We have seen that when colliding waves combine (add) the result can either be bigger or smaller than the original waves.
  • We say the waves add “constructively” or “destructively” depending on the relative sign of each wave.
  • In general, we will have both happening

see Figure 16.8

superposition interference10
Superposition & Interference
  • Consider two harmonic waves A and B meeting.
    • Same frequency and amplitudes, but phases differ.
  • The displacement versus time for each is shown below:

A(t)

B(t)

What does C(t) =A(t)+B(t) look like ??

superposition interference11
Superposition & Interference
  • Add the two curves,
    • A = A0 cos(kx – wt)
    • B = A0 cos (kx – wt - f)
  • Easy,
    • C = A + B
    • C = A0 (cos(kx – wt) + cos (kx – wt + f))
    • formula cos(a)+cos(b) = 2 cos[ 1/2(a+b)] cos[1/2(a-b)]
    • Doing the algebra gives,
    • C = 2 A0 cos(f/2) cos(kx – wt - f/2)
superposition interference12
Superposition & Interference
  • Consider,
    • C = 2 A0 cos(f/2) cos(kx – wt - f/2)

A(t)

B(t)

Amp = 2 A0 cos(f/2)

C(kx-wt)

Phase shift = f/2

lecture 35 act 1 superposition
Lecture 35, Act 1Superposition
  • You have two continuous harmonic waves with the same frequency and amplitude but a phase difference of 170° meet. Which of the following best represents the resultant wave?

Original wave

(other has different phase)

B)

A)

D)

C)

E)

lecture 35 act 1 superposition14

D)

Lecture 35, Act 1Superposition
  • The equation for adding two waves with different frequencies, C = 2 A0 cos(f/2) cos(kx – wt - f/2).
  • The wavelength (2p/k) does not change.
  • The amplitude becomes 2Aocos(f/2). Withf=170, we have cos(85°) which is very small, but not quite zero.
  • Our choice has same l as original, but small amplitude.
wave power

See text: 16.8

Wave Power
  • A wave propagates because each part of the medium communicates its motion to adjacent parts.
    • Energy is transferred since work is done !
  • How much energy is moving down the string per unit time. (i.e. how much power ?)

P

wave power16

See text: 16.8

Wave Power...
  • Think about grabbing the left side of the string and pulling it up and down in the y direction.
  • You are clearly doing work since F.dr > 0 as your hand moves up and down.
  • This energy must be moving away from your hand (to the right) since the kinetic energy (motion) of the string stays the same.

P

how is the energy moving

x

Power P = F.v

x

F

v

See text: 16.8

How is the energy moving?
  • Consider any position x on the string. The string to the left of x does work on the string to the right of x, just as your hand did:

see Figure 16-15

power along the string

y

x

vy

F

v

dy

dx

Recall

sin  

cos  1

for small 

tan  

See text: 16.8

Power along the string.
  • Since v is along the y axis only, to evaluate Power = F.v we only need to find Fy = -Fsin -F  if  is small.
  • We can easily figure out both the velocity v and the angle  at any point on the string:
  • If
average power

We are often just interested in the average power movingdown the string. To find this we recall that the averagevalue of the function sin2(kx - t) is 1/2 and find that:

See text: 16.8

Average Power
  • We just found that the power flowing past location x on the string at time t is given by:
  • It is generally true that wave power is proportional to thespeed of the wave v and its amplitude squared A2.
recap useful formulas
Recap & Useful Formulas:

y

A

x

  • General harmonic waves
  • Waves on a string

tension

mass / length

lecture 35 act 2 wave power
Lecture 35, Act 2Wave Power
  • A wave propagates on a string. If both the amplitude and the wavelength are doubled, by what factor will the average power carried by the wave change ?

i.e. Pfinal/Pinit = X

(a) 1/4 (b) 1/2 (c) 1 (d) 2 (e) 4

initial

final

waves wavefronts and rays

3-D Representation

RAYS

Wave Fronts

Waves, Wavefronts, and Rays
  • Up to now we have only considered waves in 1-D but we live in a 3-D world.
  • The 1-D equations are applicable for a 3-D plane wave.
  • A plane wave travels in the +x direction (for example) and has no dependence on y or z,
waves wavefronts and rays24

3d representation

Shading represents

density

wave fronts

rays

Waves, Wavefronts, and Rays
  • Sound radiates away from a source in all directions.
  • A small source of sound produces a spherical wave.
  • Note any sound source is small if you are far enough away from it.
waves wavefronts and rays25
Waves, Wavefronts, and Rays
  • Note that a small portion of a spherical wave front is well represented as a plane wave.
waves wavefronts and rays26
Waves, Wavefronts, and Rays
  • If the power output of a source is constant, the total power of any wave front is constant.
  • The Intensity at any point depends on the type of wave.
lecture 35 act 3 spherical waves
Lecture 35, Act 3Spherical Waves
  • You are standing 10 m away from a very loud, small speaker. The noise hurts your ears. In order to reduce the intensity to 1/2 its original value, how far away do you need to stand?

(a) 14 m (b) 20 m (c) 30 m (d) 40 m

lecture 35 act 4 traveling waves
Lecture 35, Act 4Traveling Waves

Two ropes are spliced together as shown.

A short time after the incident pulse shown in the diagram reaches the splice, the ropes appearance will be that in

  • Can you determine the relative amplitudes of the transmitted and reflected waves ?
lecture 35 act 3b plane waves
Lecture 35, Act 3bPlane Waves
  • You are standing 0.5 m away from a very large wall hanging speaker. The noise hurts your ears. In order to reduce the intensity you walk back to 1 m away. What is the ratio of the new sound intensity to the original?

(a) 1 (b) 1/2 (c) 1/4 (d) 1/8

speaker

1 m

recap of today s lecture
Recap of today’s lecture
  • Chapter 16
    • Waves on a string
    • Superposition
    • Power