First, some interesting sites related to today's discussion:
http://library.thinkquest.org/19537/cgi-bin/showharm.cgi
This site contains a program that allows you to put together harmonics and see what they sound like together.
http://www.pbs.org/wgbh/nova/bridge/meetsusp.html#clips
This site describes one of the most catastrophic applications of resonance, the Tacoma Narrows Bridge collapse.
Now, some solutions:
33. Since the nodes are now at half the distance, and the wave velocity is the same, the frequency must double to 8 Hz. This is beyond the highest frequency of earthquakes.
36. The tube must be closed at one end - look at the ratio of the frequencies to each other, as they are odd fractions. The fundamental frequency is 88 Hz.
37. f = 5v/2L, so v = 247.5 m/s.
38. Open at both ends, 7v/2L = 33o Hz, L = 3.606 m
75. n*lambda/4 = L1, (n+2)*lambda/4 = L2. Subtract the two equations, and you get that lambda = 2(L2 - L1), leaving that f = 630 Hz if the speed of sound is 340 m/s.
Monday, March 10, 2008
Friday, March 7, 2008
Standing Waves
Q25. The nodes are locations of zero amplitude, so these positions could be touched without disturbing the wave.
Problems:
18. amplitude = 0.3500 meters, frequency = .8754 Hz, period = 1.142 seconds, total energy = .7411 J, KE = .6807 J and PE = .0604 J
51. 440 Hz, 880 Hz, 1320 Hz, 1760 Hz (fn = nv/2L)
53. 70 Hz, 140 Hz, 210 Hz, 350 Hz
55. nodes are always a half wavelength apart from each other - wavelength is 19.37 cm, so nodes are half of this, 9.685 cm.
56. 87.5 Hz, n = 3 and 4 for the given frequencies.
60. m = 4f2L2*mu/(n2g), so m = 1.421 kg, .355 kg, .0568 kg for (a), (b), and (c).
Web sites from today:
http://www.colorado.edu/physics/2000/applets/fourier.html- Constructive/Destructive Interference
http://home.austin.rr.com/jmjensen/JeffString.html
Reflected Waves
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=19 - Waves traveling in opposite directions making standing waves
Standing Wave - nodes and antinodes
http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html - Vibrating string, standing waves fixed at both ends.
http://www.walter-fendt.de/ph11e/stlwaves.htm - open/closed Tube with standing waves
Thursday, March 6, 2008
Wave Motion Solutions
Q10 . The water sloshing back and forth has a specific frequency associated with it because of the size of the pan and the speed of the waves on the surface of the water. The only way to get the water to slosh is if the swinging occurs at this special frequency.
Q18 (Not 8, I'm sorry!). Hitting across the end creates a transverse wave, hitting perpendicular to the end creates a longitudinal wave.
Problems.
34. 2.83 m/s
35. 1.259 m
39. 0.332 s
41. The speed of sound in water is 1440 m/s. The sound must travel down to the ocean floor and back. The distance is therefore 2160 meters.
Q18 (Not 8, I'm sorry!). Hitting across the end creates a transverse wave, hitting perpendicular to the end creates a longitudinal wave.
Problems.
34. 2.83 m/s
35. 1.259 m
39. 0.332 s
41. The speed of sound in water is 1440 m/s. The sound must travel down to the ocean floor and back. The distance is therefore 2160 meters.
Wednesday, March 5, 2008
EMI Problem Answers
1986B4.
a) 3 V
b) clockwise
c) 0.6 N
d) 1.8 W
1982B5.
a) 0.06 Webers
b) 0.06 V
c) -0.3 A from t = 0 to t = 2 seconds, 0 A from t = 2 to t = 4 seconds, and +0.15 A from t = 4 to t = 6 seconds.
3.
a) final speed is (2gy0)^(1/2)
b) I = Bh/R*(2gy0)^(1/2)
c) flux is increasing into the page, so induced current increases the flux out of the page. Current is CCW.
d) For flux:
Flux increases linearly to whB at w; Stays constant until 3w, and then decreases to zero at 4w.
For current: The magnitude of I is the value given in part (b).
Positive I from 0 to w, 0 from w to 3w, -I from 3w to 4w, and then zero from 4w to 5w.
4.
a) m = ILB/g = B^2L^2v/gR = 0.143 kg
b) 1.51 J
c) 1.51 J
5.
a) 0.3 Volts
b) 0.06 A, counter clockwise
c) 0.018 W
d) F = 0.036 N
e) 20 more turns means potential difference increases by a factor of 20. This also means twice the resistance of the wire. Since current I = deltaV/R, the factors of 20 will divide out, leaving the current the SAME.
a) 3 V
b) clockwise
c) 0.6 N
d) 1.8 W
1982B5.
a) 0.06 Webers
b) 0.06 V
c) -0.3 A from t = 0 to t = 2 seconds, 0 A from t = 2 to t = 4 seconds, and +0.15 A from t = 4 to t = 6 seconds.
3.
a) final speed is (2gy0)^(1/2)
b) I = Bh/R*(2gy0)^(1/2)
c) flux is increasing into the page, so induced current increases the flux out of the page. Current is CCW.
d) For flux:
Flux increases linearly to whB at w; Stays constant until 3w, and then decreases to zero at 4w.
For current: The magnitude of I is the value given in part (b).
Positive I from 0 to w, 0 from w to 3w, -I from 3w to 4w, and then zero from 4w to 5w.
4.
a) m = ILB/g = B^2L^2v/gR = 0.143 kg
b) 1.51 J
c) 1.51 J
5.
a) 0.3 Volts
b) 0.06 A, counter clockwise
c) 0.018 W
d) F = 0.036 N
e) 20 more turns means potential difference increases by a factor of 20. This also means twice the resistance of the wire. Since current I = deltaV/R, the factors of 20 will divide out, leaving the current the SAME.
Tuesday, March 4, 2008
Lenz's Law Solutions
I apologize for leaving it out, but please use the image below with Problem 2 from the handout from today's lesson:
Questions:
15. The change in flux causes an induced current within the aluminum. The magnetic field created by the induced current is attracted to the magnetic field of the bar magnet, opposing the force tending to pull the sheet out of the magnet. This induced magnetic field forms because of the induced current, not because the magnetic properties of aluminum.
16. We discussed this in class!
Problems:
3. Counter clock-wise
4. Field lines going out from magnet to the right. Moving it through loop increases the flux going into loop towards the right. By Lenz' law, the induced current will increase the magnetic flux going into the loop towards the left. This will create a current that flows clockwise through the loop, and from right to left through the resistor.
7.
a)The current through the outer loop decreases since the resistance increases. This means that the magnetic field (and therefore the flux) through the inner loop out of the page is decreasing. By Lenz's law, the induced current will attempt to increase the flux out of the page to oppose the change. This will come from a counter-clockwise current.
b) If the loop was moved outside to the left, the flux would be decreasing into the page through the loop. The induced current would try to increase flux into the page through the loop, which would result from a clockwise current.
8.
a) counter-clockwise
b) clockwise
c) zero (no change in flux)
d) counter-clockwise
Questions:
15. The change in flux causes an induced current within the aluminum. The magnetic field created by the induced current is attracted to the magnetic field of the bar magnet, opposing the force tending to pull the sheet out of the magnet. This induced magnetic field forms because of the induced current, not because the magnetic properties of aluminum.
16. We discussed this in class!
Problems:
3. Counter clock-wise
4. Field lines going out from magnet to the right. Moving it through loop increases the flux going into loop towards the right. By Lenz' law, the induced current will increase the magnetic flux going into the loop towards the left. This will create a current that flows clockwise through the loop, and from right to left through the resistor.
7.
a)The current through the outer loop decreases since the resistance increases. This means that the magnetic field (and therefore the flux) through the inner loop out of the page is decreasing. By Lenz's law, the induced current will attempt to increase the flux out of the page to oppose the change. This will come from a counter-clockwise current.
b) If the loop was moved outside to the left, the flux would be decreasing into the page through the loop. The induced current would try to increase flux into the page through the loop, which would result from a clockwise current.
8.
a) counter-clockwise
b) clockwise
c) zero (no change in flux)
d) counter-clockwise
Monday, March 3, 2008
Faraday's Law & Electromagnetic Induction
Question 2:
Magnetic field describes how the influence of magnetic force is transmitted through space. Magnetic force describes how magnetic flux passes through a specific region of space, specifically a loop or area through which magnetic field lines travel. It is possible to have magnetic field and not have magnetic flux (if area = 0 or if the angle between B and the normal vector is 90 degrees), but it is not possible to have magnetic flux without magnetic field.
Problems:
2. 0.147 Volts
10.
a) 0.169 Volts
12.
b) 4.29 x 10-2 Volts
c) 1.72 mA
13.
a) .841 m/s
b) .757 N/C
89. 7.27 x 10-3 J
Here are a couple of sites with fun EMI tools to play with:
http://www.ngsir.netfirms.com/englishhtm/Induction.htm
http://higheredbcs.wiley.com/legacy/college/halliday/0471320005/simulations6e/index.htm?newwindow=true
Magnetic field describes how the influence of magnetic force is transmitted through space. Magnetic force describes how magnetic flux passes through a specific region of space, specifically a loop or area through which magnetic field lines travel. It is possible to have magnetic field and not have magnetic flux (if area = 0 or if the angle between B and the normal vector is 90 degrees), but it is not possible to have magnetic flux without magnetic field.
Problems:
2. 0.147 Volts
10.
a) 0.169 Volts
12.
b) 4.29 x 10-2 Volts
c) 1.72 mA
13.
a) .841 m/s
b) .757 N/C
89. 7.27 x 10-3 J
Here are a couple of sites with fun EMI tools to play with:
http://www.ngsir.netfirms.com/englishhtm/Induction.htm
http://higheredbcs.wiley.com/legacy/college/halliday/0471320005/simulations6e/index.htm?newwindow=true
Sunday, March 2, 2008
Magnetic Field from a Wire/Magnetic Force
1.
a) 5 x 10-5T
b) West
c) Net B-field is of magnitude 7.07 x 10-5T directed at 45 degrees North of West. (It is not enough to say 7.07 x 10-5T at 45 degrees!)
2. a = 0.0076 m/s^2
3. 48 A
4.
a) negative
b) Electric field is directed towards the top plate.
c) 2280 V
d) 9.5 x 106 C/kg
5.
a) perpendicular to the plane of the page, directed OUT of the page.
b) 2 x 10-7T
c) negative
d) 5 x 10-7C
e) 0.2 N/C directed to the left
6.
a) perpendicular to the page, out of the page
b) 1.9 x 10-15N
c) 19 cm
d) 1.2 x 104N/C
e) The electric field is in the plane of the page, directed towards the top of the page.
7. This problem is very similar to #55 from yesterday. Important points: What is R in terms of given quantities? If the charge q crosses a potential difference V, what is the final speed if it starts from rest?
Additionally, some help on #67:
Many of you are finding the speed of the electrons, and are then able to figure out that you need to find the radius of curvature of the path. The image below suggests a way to find the deflection of the electrons from their intended path.
a) 5 x 10-5T
b) West
c) Net B-field is of magnitude 7.07 x 10-5T directed at 45 degrees North of West. (It is not enough to say 7.07 x 10-5T at 45 degrees!)
2. a = 0.0076 m/s^2
3. 48 A
4.
a) negative
b) Electric field is directed towards the top plate.
c) 2280 V
d) 9.5 x 106 C/kg
5.
a) perpendicular to the plane of the page, directed OUT of the page.
b) 2 x 10-7T
c) negative
d) 5 x 10-7C
e) 0.2 N/C directed to the left
6.
a) perpendicular to the page, out of the page
b) 1.9 x 10-15N
c) 19 cm
d) 1.2 x 104N/C
e) The electric field is in the plane of the page, directed towards the top of the page.
7. This problem is very similar to #55 from yesterday. Important points: What is R in terms of given quantities? If the charge q crosses a potential difference V, what is the final speed if it starts from rest?
Additionally, some help on #67:
Many of you are finding the speed of the electrons, and are then able to figure out that you need to find the radius of curvature of the path. The image below suggests a way to find the deflection of the electrons from their intended path.
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