WEB
Final Exam: smartPhysics units 01-24 Day: Tue. Dec. 15, 2015 Time: 3:30-5:30 pm (both sections)
Location Section 01: WEB L104 Section 10: WEB L101 Practice problems for remaining units (21-24) not yet covered in practice problem sheets have been posted on CANVAS and on the class web page http://www.physics.utah.edu/~woolf/2210_Jui/rev5.pdf ***Review for Units 21-24 Friday Dec 11, 1pm-2pm JFB 101
At least ONE, and up to TWO of the six problems on the final exam will be on the materials from units 21-24.
Harmonic waves
If we choose π(π₯) = π΄ cosππ₯ as our function to propagate in the +x direction then the result is a harmonic wave:
π¦ π₯, π‘ = π΄ cosπ π₯ β π£π‘ = π΄ cos ππ₯ β ππ‘ Where we have made the substitution π = ππ£ β’ With time fixed: the wave advances ONE cycle over length given by πΞπ₯ = 2π. But this
is the definition of wavelength, π: i.e. ππ = 2π. So we have
π =2ππ , π =
2ππ
π is called the βwave numberβ and has units mβ1 or rad/m. β’ Pick ONE value of π₯ on the string: ONE cycle of the oscillation = one period: Οπ = 2π:
π =2ππ , π =
2ππ , π =
1π =
π2π , π = 2ππ
And again we refer to π as the βangular frequencyβ
π΄cosπ π β ππ
π
Poll 12-07-02
We have shown that the functional form y(x,t) = Acos(kx-Οt) represents a wave moving in the +x direction. Which of the following represents a wave moving in the -x direction? A. Acos(Οt-kx) B. Acos(kx-Οt) C. Acos(kx+Οt)
Generic Solution to the Equation of Motion Equation of Motion (π£ = π πβ for a string under tension)
π2π¦ππ₯2
=1π£2π2π¦ππ‘2
Has generic solutions of the form π¦1 π₯, π‘ = π π₯ β π£π‘ , π¦2 π₯, π‘ = π π₯ + π£π‘
Where π π₯ is any single-variable function, for example (π = 0.3 in the plot)
π π₯ =π
π₯2 + π2 β¦ππππ ππ ππππ πππ ππΈπ΄πΈπΈππ
A translational transformation (or just βtranslationβ) of π π₯ β π , i.e.
π π₯ β π =π
π₯ β π 2 + π2 β¦ππππ ππ ππππ πππ ππΈπ΄πΈπΈππ
Shifts the function to the right/left (+/-x dir.) by π for π>0, and c<0, respectively
π π π π β π π π β π π π+ π
Generic Solutions to the Equation of Motion
π π₯ + π£π‘ , for example π π₯ + π£π‘ =π
π₯ + π£π‘ 2 + π2
Represents a βpulseβ of shape given by f(x) that is traveling to the right (+x direction) at speed π£: (π£>0). We will now show that π π₯ + π£π‘ is a solution ***
ππ(π₯ + π£π‘) ππ₯ = πβ² π₯ + π£π‘ β
π π₯ + π£π‘ππ₯ = πβ² π₯ + π£π‘
π2π(π₯ + π£π‘) ππ₯2 =
ππβ²(π₯ + π£π‘) ππ₯ = πβ²β² π₯ + π£π‘ β
π π₯ + π£π‘ππ₯ = πβ²β² π₯ + π£π‘
ππ(π₯ + π£π‘) ππ‘ = πβ² π₯ + π£π‘ β
π π₯π‘ππ‘ = +π£ β πβ² π₯ + π£π‘
π2π(π₯ + π£π‘) ππ‘2 = π£ β
ππβ² π₯ + π£π‘ ππ‘ = π£πβ²β² π₯ + π£π‘ β
π π₯ + π£π‘ππ‘ = π£πβ²β² π₯ + π£π‘ β π£
= π£2πβ²β² π₯ + π£π‘
β1π£2π2π(π₯ + π£π‘)
ππ‘2 =1π£2 β π£
2πβ²β² π₯ + π£π‘ = πβ²β² π₯ + π£π‘ =π2π π₯ + π£π‘
ππ₯2 β¦ π. π.π.
π π+ ππ
π
Poll 12-07-03
The pulse in Case A is described by the function y(x,t) = P(x-vt).
Which of the following functions describes the pulse in Case B?
A. y(x,t) = P(x+vt) B. y(x,t) = -P(x+vt) C. y(x,t) = -P(x-vt)
Example 23.2 A wave traveling along the x axis is described mathematically by the equation π¦ = 0.17m β cos(0.54ππ₯ + 8.2ππ‘), where π¦ is the displacement (in meters), π‘ is in seconds, and π₯ is in meters. (a) What is the speed of the wave? (b) What is the displacement of the particle at π₯ = 0.30 m at time π‘= 7.2 s?
Solution (a) (Wave speed)=(frequency)(wavelength) π£ = π Ξ» and we know the wave as the form π¦ = π΄ sin (ππ₯ + ππ‘ ) ** note the + sign here means the wave is traveling in the βx direction.
so Ο = 2ππ = 8.2π sβ1 f = 8.2 s-1 / 2 = 4.1 s-1 and k = 2π/οΏ½ = 0.54Ο m-1 Ξ» = 2 / (0.54 m-1) = 3.7 m π£ = π = (4.1 s-1)(3.7 m) π£ = 15 m/s (b)
π¦ 0.30 m, 7.2s = 0.17 m cos 0.54ππ₯ + 8.2ππ‘= 0.17 m cos + (0.54Ο mβ1)(0.30 m) + (8.2Ο sβ1)(7.2 s)= 0.17 π cos 185.9885683 rad
cos(185.9885683 rad)= β0.805 π¦(0.30 m, 7.2s) = -0.137 m
Note: The phase angles ππ₯ β ππ‘ or ππ₯ + ππ‘ are measured in radians, not degrees. SET YOUR CALUCLATOR IN RADIAN (RAD) MODE WHEN WORKING ON THIS SORT OF PROBLEM Also: Keep as many digits as you get from the calculator for the phase angleβ¦itβs important to NOT round off before the final result you want
When the pulses merge, the Slinky assumes a shape that is the sum of the shapes of the individual pulses.
In this case, the pulses (both upward) reinforce one another and we have βconstructive interferenceβ
Linear Superposition and Interference
http://www.ablongman.com/mullin/AnimaImages/ConsInterf.gif
When the pulses merge, the Slinky assumes a shape that is the sum of the shapes of the individual pulses.
In this case, the pulses (one up, one down) work against one another, and cancel each other at one instant. Here we have βdestructive interferenceβ
http://www.ablongman.com/mullin/AnimaImages/DesIntef.gif
Constructive and Destructive interference Video
http://www.youtube.com/watch?v=P_rK66GFeI4
Two waves with the same amplitude and frequency traveling in opposite directions on the SAME medium interfere in such a way as to produce a stationary oscillating pattern This is called a Standing Wave (e.g. the line segment between the two bobs in the ripple tank)
yAgain: the locations where you always have destructive interference are called βnodesβ. The locations with maximum constructive interference are βanti-nodesβ
[ ])/22sin()/22sin(),(),(),()/22sin(),( ),/22sin(),(
Ξ»ΟΟΞ»ΟΟΞ»ΟΟΞ»ΟΟ
xftxftAtxytxytxyxftAtxyxftAtxy
++β=+=+=β=
β+
β+
The math of Standing Waves: the rightward and leftward waves are given, respectively:
We again use the identity:
[ ] )2sin()/2cos(2),()2sin()/2cos(2),(ftxAtxy
ftxAtxyΟΞ»Ο
ΟΞ»Ο=β
β=
http://www.kettering.edu/physics/drussell/Demos/superposition/super3.gif
Transverse Standing Waves It is possible to produce self-sustaining standing waves on nearly ideal (very little energy loss as the wave propagates) by taking advantage of reflections There are two special cases of reflections: (1) Reflection from a fixed end (where y=0 always)
Click to start movie http://www.youtube.com/watch?v=LTWHxZ6Jvjs
http://www.kettering.edu/physics/drussell/Demos/reflect/hard.gif
Transverse Standing Waves By providing small impulses at one end, one can generate a large standing wave from the reflections (the hand in this case is very nearly a node), provided that the frequency is right , such that the boundary conditions are satisfied Boundary condition: Two fixed ends (all string musical instruments are made this way)
When you get the right frequencies where small stimuli maintains a large standing wave: we have what is called resonance . All musical instruments depends on resonance to generate sound of a definite pitch/frequency.
http://www.youtube.com/watch?v=jovIXzvFOXo
17.5 Transverse Standing Waves
,4,3,2,1 2
=
= n
LvnfnString fixed at both ends 17
f1 f2=2 f1 f3=3 f1
First Harmonic n = 1
Second Harmonic n = 2
Third Harmonic n = 3
==β=
LvvfL
21
2 11
1
λλ
==β=
LvvfL
22
22
12
2
λλ
==β=
LvvfL
23
23
13
3
λλ
L=length of string The progression of harmonics adds ONE node at a time
http://www.physicsclassroom.com/class/waves/u10l4eani1.gif
Poll 12-09-01 Suppose the strings on your guitar are 24β long as shown.
The frets are the places along the neck where you can put your finger to make the wavelength shorter, and appear as horizontal white lines on the picture.
When no frets are being pushed the frequency of the highest string is 4 times higher than the frequency of the lowest string.
Is it possible to play the lowest string with your finger on any of the frets shown and hear the same frequency as the highest string?
A. Yes B. No
People have discovered that taking videos with iphone cameras leads to interesting effects https://www.youtube.com/watch?v=TKF6nFzpHBU Some claim this is seeing the actual waves traveling down the strings Wellβ¦. Not quite These cameras use CMOS sensors and βrolling shuttersβ
Same type of camera alos takes other really fun videos https://www.youtube.com/watch?v=eTW0rNgMcKk This is a type of βaliasingβ which is a big area of study in signal processing See http://nofilmschool.com/2014/02/how-rolling-shutter-affects-our-perceptions-of-the-andromeda-galaxy For an fun lesson on this topic
Top Related