Photoelectric Effect
http://www.lon-capa.org/~mmp/kap28/PhotoEffect/photo.htm
More visual Photoelectric effect:
http://phet.colorado.edu/simulations/photoelectric/photoelectric.jnlp
Solutions:
2000B5
a)
i) 4.5 eV or 7.2 x 10-19 J
ii) 1.26 x 106 m/s
b) 183 nm
c) 5.56 x 1014 Hz
1980B3
a) The graph should be straight forward, but you SHOULD know that all four points cannot be on the same line!
b) 0.75 x 1014 Hz
c) 3.1 eV
d) Energy of electrons is 5 eV according to data, so the stopping potential will be 5 volts.
e) Charges move in circular paths when they are in a magnetic field perpendicular to their velocity. This is a possible reason for this behavior.
Page 855:
16. 4.14 x 1014 Hz
17. 401 nm
19. The shortest visible wavelength photons have energies of 3.1 eV (400 nm wavelength). Copper and Iron have work functions that are greater than this, so they will not eject photo electrons in visible light.
20.
a) 3.49 x 10-19 J or 2.179 eV.
b) 0.930 V
Page 885:
P32. 4.14 x 10-11 m (Don't worry about the longest wavelength.)
P33. 41,000 V
Tuesday, March 31, 2009
Monday, March 30, 2009
From Handout:
1.
a) 2.5 mm
b) wavelength = 385 nm, frequency = 6 x 1014 Hz
c) Since the wavelength decreases, and the spacing between maxima is proportional to wavelength, the space between the maxima will decrease.
1991B6.
a) 3.9 x 10-5 m
b) 9.6 x 10-3 m
Problems:
7. 8.5 x 10-5m
10. The important thing here is to notice that the ratio d/L is the same for both. You should obtain the equation that 2*lambda_1/y = m*lambda_2/2*y, with y the same for both since the question asks for the wavelength at which a minimum appears at the same location. Solving for a wavelength lambda_2 in the visible spectrum, lambda_2 = 613 nm
20. Solve for theta using sin(theta)= m*lambda/W, and then use the fact that L*tan(theta) = height above the center. The distance is 1.591 meters.
21. Notice that this says the distance between maxima, not minima. The path difference must be half a wavelength more than that required for the minimum. The total angle is 19.5 degrees across the central maximum, so the angle between the central maximum and the first minimum is 9.75 degrees. sin(9.75) = 1.5*lambda/W, so W = 5606 nm.
1.
a) 2.5 mm
b) wavelength = 385 nm, frequency = 6 x 1014 Hz
c) Since the wavelength decreases, and the spacing between maxima is proportional to wavelength, the space between the maxima will decrease.
1991B6.
a) 3.9 x 10-5 m
b) 9.6 x 10-3 m
Problems:
7. 8.5 x 10-5m
10. The important thing here is to notice that the ratio d/L is the same for both. You should obtain the equation that 2*lambda_1/y = m*lambda_2/2*y, with y the same for both since the question asks for the wavelength at which a minimum appears at the same location. Solving for a wavelength lambda_2 in the visible spectrum, lambda_2 = 613 nm
20. Solve for theta using sin(theta)= m*lambda/W, and then use the fact that L*tan(theta) = height above the center. The distance is 1.591 meters.
21. Notice that this says the distance between maxima, not minima. The path difference must be half a wavelength more than that required for the minimum. The total angle is 19.5 degrees across the central maximum, so the angle between the central maximum and the first minimum is 9.75 degrees. sin(9.75) = 1.5*lambda/W, so W = 5606 nm.
Friday, March 27, 2009
There's this thin film on my teeth...(Eww)
Hi everyone,
Here is the link to the physics lesson website - be sure to check out the lesson on thin film interference, as well as on diffraction.
Please also be aware that there will be construction Saturday on the 6 line downtown on the way to the REACH workshop at Baruch College. Please give yourself plenty of time to get there by 8:00 AM.
From handout:
4. thickness = 1.05 x 10-7 m
5. thickness = 96.1 nanometers
1984B5
a) 5 x 1014 Hz
b) 4.8 x 10-7 m
c) 1.2 x 10-7 m
d) 2.4 x 10-7 m
2000B5.
I apologize for not getting this solutions up quickly, but you must be aware that part (a) of this question is worth 8 of the 15 points. You need to make sure you draw BOTH angles of reflection, BOTH angles of refraction (after using Snell's Law), and include the angle of incidence (35 degrees) of the light ray upon hitting the bottom surface of the glass.
Parts b and c are, by comparison, relatively tame. I will post this ASAP.
38. wavelength = 643 nm, which is colored red.
40. 169 nm
43. 27*lambda/n = 2t, so t = 9045 nanometers, or 9.045 x 10-6 m
45. 2t = n*(640 nm/1.36), and 2t = (n + 1/2)*512 nm/1.36. Solving the system, t = 471 nanometers.
Here is the link to the physics lesson website - be sure to check out the lesson on thin film interference, as well as on diffraction.
Please also be aware that there will be construction Saturday on the 6 line downtown on the way to the REACH workshop at Baruch College. Please give yourself plenty of time to get there by 8:00 AM.
From handout:
4. thickness = 1.05 x 10-7 m
5. thickness = 96.1 nanometers
1984B5
a) 5 x 1014 Hz
b) 4.8 x 10-7 m
c) 1.2 x 10-7 m
d) 2.4 x 10-7 m
2000B5.
I apologize for not getting this solutions up quickly, but you must be aware that part (a) of this question is worth 8 of the 15 points. You need to make sure you draw BOTH angles of reflection, BOTH angles of refraction (after using Snell's Law), and include the angle of incidence (35 degrees) of the light ray upon hitting the bottom surface of the glass.
Parts b and c are, by comparison, relatively tame. I will post this ASAP.
38. wavelength = 643 nm, which is colored red.
40. 169 nm
43. 27*lambda/n = 2t, so t = 9045 nanometers, or 9.045 x 10-6 m
45. 2t = n*(640 nm/1.36), and 2t = (n + 1/2)*512 nm/1.36. Solving the system, t = 471 nanometers.
Wednesday, March 25, 2009
It's Critical that you get the right angle!
Solutions to the AP Problems:
79B6.
a) Remember that you MUST use the angles relative to the normal vector - n1 sin(t1) = n2 sin (t2)
n2 = 1.327
n = c/v so v = 2.26 x 108 m/s
b) If you try to use Snell's law at point Q, you will find you end up trying to take the inverse sine of a number greater than 1 - not possible. This means the ray will NOT refract, but instead reflect. This is total internal reflection. You can justify the fact that it reflects either by calculating the critical angle and showing that the 53 degrees is greater than the critical angle (49 degrees), or by showing that when refraction does not occur, reflection does.
c) If the lens was made of plastic in air, it would be a converging lens. Since it is the reverse, it will be a diverging lens, and the rays will spread out.
88B5.
a) The ray will be refracted slightly downwards from the horizontal dotted line.
b)
The ray will pass directly through the left interface because the incident angle is 0. At the right interface, the angle from the normal is 37 degrees. By Snell's law, the refracted ray will have an angle of 64.5 degrees.
Be careful though, as the question asks for the angle from the horizontal. To get this, you must subtract the 37 degrees from the 64.5 degrees, giving a final answer of 27.5 degrees.
c) Here you must assume that the incident angle of 37 degrees is the critical angle.
By Snell's law, theta_1 = 37, n2 = 1.0 (for air), and theta_2 = 90 since we are assuming total internal reflection occurs. Solving, n1 = 1.667
d) Here the light ray will again be refracted below the horizontal line, but the angle should not be as great as it was for part (a).
e) Using Snell's law again with n1 = 1.667, theta_1 = 37 degrees, and n2 = 1.33, we can solve for theta_2 = 49.0 degrees. You must again subtract 37 degrees from this value to obtain the angle from the horizontal of 12 degrees below.
93B4.
a) v = c/n = 1.875 x 108 m/s
b) There are two ways to do this - the fast way is to remember the equation from Monday that the wavelength in the material lambda' = lambda/n where lambda is the wavelength in a vacuum. 700 nm/1.5 = 467 nm.
We can also use c = f*lambda to find the frequency of the red light in a vacuum:
f = 3E8 m/s/(700 E-9 m) = 4.28 x 1014 Hz
The frequency stays the same when a wave changes from one material to another, so this is also the frequency in the glass. We also know the speed of red light in the glass from part a.
lambda = v/f = 1.875 x 108 m/s/(4.28 x 1014 Hz) = 438 nm (either answer would be acceptable)
c) Frequency can be found the same as before - make SURE before you use the v = f*lambda equation, you are using the speed and wavelength for the same medium!
f = 3E8 m/s/(700 E-9 m) = 4.28 x 1014 Hz
d) The key here is to notice that the incident rays have the SAME initial angle hitting the right side of the prism (30 degrees), as well as the same n2 = 1 (for air.) The difference is in n1.
From Snell's Law:
n1 sin(theta_1) = n2*sin(theta_2)
sin(theta_2) = n1*sin(theta_1)
Since n1 is greater for the blue light, theta_2 will be greater for the blue light. This means that the blue light will bend further away relative to the normal than the red.
e) In this case, it is a similar situation when the rays reach the right surface (the rays again do not refract at the left surface because the incident angle is zero.) The issue here, however, is that n1 is the same for both (n1 = 1), and theta_1 is again 30 degrees.
Snell's law again:
n1 sin(theta_1) = n2*sin(theta_2)
sin(theta_2) = 1*sin(theta_1)/n1
In this case, since n1 is greater for blue light, it will refract less
than with the red light. Thus, the blue light will bend above the red light, with both rays bending towards the normal.
#3.
b) Using the thin lens equation, 1/di = 1/f - 1/do, so di = 30 cm. This should roughly match your answer for part (a).
c) hi/ho = -di/do so hi = 2*5cm = 10 cm
d) This can be a tricky one. If you use what we came up with today, you will find that the image from the first lens is on the right side of the second lens. You can use the rules as we always do, except that the primary focus of the second lens is now on the same side as the object, so the lens will act like a diverging lens.
The slick way to get this is to notice that Ray 2 leaves the first lens parallel to the axis when it enters the second lens. This ray enters lens 2 parallel, so it is brought to the primary focus of lens 2. Ray 1, after passing through lens 1, passes through the center of lens 2, which means it will pass through undeflected. This will result in these two rays converging at a single point to the right of lens 2. This is the location of the image, roughly 7 cm to the right of lens 2.
e) The image is inverted and smaller than the original object.
79B6.
a) Remember that you MUST use the angles relative to the normal vector - n1 sin(t1) = n2 sin (t2)
n2 = 1.327
n = c/v so v = 2.26 x 108 m/s
b) If you try to use Snell's law at point Q, you will find you end up trying to take the inverse sine of a number greater than 1 - not possible. This means the ray will NOT refract, but instead reflect. This is total internal reflection. You can justify the fact that it reflects either by calculating the critical angle and showing that the 53 degrees is greater than the critical angle (49 degrees), or by showing that when refraction does not occur, reflection does.
c) If the lens was made of plastic in air, it would be a converging lens. Since it is the reverse, it will be a diverging lens, and the rays will spread out.
88B5.
a) The ray will be refracted slightly downwards from the horizontal dotted line.
b)
The ray will pass directly through the left interface because the incident angle is 0. At the right interface, the angle from the normal is 37 degrees. By Snell's law, the refracted ray will have an angle of 64.5 degrees.
Be careful though, as the question asks for the angle from the horizontal. To get this, you must subtract the 37 degrees from the 64.5 degrees, giving a final answer of 27.5 degrees.
c) Here you must assume that the incident angle of 37 degrees is the critical angle.
By Snell's law, theta_1 = 37, n2 = 1.0 (for air), and theta_2 = 90 since we are assuming total internal reflection occurs. Solving, n1 = 1.667
d) Here the light ray will again be refracted below the horizontal line, but the angle should not be as great as it was for part (a).
e) Using Snell's law again with n1 = 1.667, theta_1 = 37 degrees, and n2 = 1.33, we can solve for theta_2 = 49.0 degrees. You must again subtract 37 degrees from this value to obtain the angle from the horizontal of 12 degrees below.
93B4.
a) v = c/n = 1.875 x 108 m/s
b) There are two ways to do this - the fast way is to remember the equation from Monday that the wavelength in the material lambda' = lambda/n where lambda is the wavelength in a vacuum. 700 nm/1.5 = 467 nm.
We can also use c = f*lambda to find the frequency of the red light in a vacuum:
f = 3E8 m/s/(700 E-9 m) = 4.28 x 1014 Hz
The frequency stays the same when a wave changes from one material to another, so this is also the frequency in the glass. We also know the speed of red light in the glass from part a.
lambda = v/f = 1.875 x 108 m/s/(4.28 x 1014 Hz) = 438 nm (either answer would be acceptable)
c) Frequency can be found the same as before - make SURE before you use the v = f*lambda equation, you are using the speed and wavelength for the same medium!
f = 3E8 m/s/(700 E-9 m) = 4.28 x 1014 Hz
d) The key here is to notice that the incident rays have the SAME initial angle hitting the right side of the prism (30 degrees), as well as the same n2 = 1 (for air.) The difference is in n1.
From Snell's Law:
n1 sin(theta_1) = n2*sin(theta_2)
sin(theta_2) = n1*sin(theta_1)
Since n1 is greater for the blue light, theta_2 will be greater for the blue light. This means that the blue light will bend further away relative to the normal than the red.
e) In this case, it is a similar situation when the rays reach the right surface (the rays again do not refract at the left surface because the incident angle is zero.) The issue here, however, is that n1 is the same for both (n1 = 1), and theta_1 is again 30 degrees.
Snell's law again:
n1 sin(theta_1) = n2*sin(theta_2)
sin(theta_2) = 1*sin(theta_1)/n1
In this case, since n1 is greater for blue light, it will refract less
than with the red light. Thus, the blue light will bend above the red light, with both rays bending towards the normal.
#3.
b) Using the thin lens equation, 1/di = 1/f - 1/do, so di = 30 cm. This should roughly match your answer for part (a).
c) hi/ho = -di/do so hi = 2*5cm = 10 cm
d) This can be a tricky one. If you use what we came up with today, you will find that the image from the first lens is on the right side of the second lens. You can use the rules as we always do, except that the primary focus of the second lens is now on the same side as the object, so the lens will act like a diverging lens.
The slick way to get this is to notice that Ray 2 leaves the first lens parallel to the axis when it enters the second lens. This ray enters lens 2 parallel, so it is brought to the primary focus of lens 2. Ray 1, after passing through lens 1, passes through the center of lens 2, which means it will pass through undeflected. This will result in these two rays converging at a single point to the right of lens 2. This is the location of the image, roughly 7 cm to the right of lens 2.
e) The image is inverted and smaller than the original object.
Tuesday, March 24, 2009
Can you Lens me a dollar? Times are tough....
Multiple Choice problems from the handout:
1. B
2. E
3. C
4. D
Questions:
20. If the object distance is very large, then by the thin lens equation, 1/d0 approaches zero. This means the image distance is equal to the focal length. This is where the film must be placed to form an image.
Problems:
48b. 390 mm
50. Since the image is on the other side of the lens, it must be a real image. This means that the image distance is positive. Using the thin lens equation, the focal length must be 39.8 cm.
52. 72 cm on the same side of the lens as the stamp (virtual image, negative image distance). Magnification = -4
60.
a) 15.2 cm, real image (opposite side of the lens as object), height is -2.78 mm
b) -12.1 cm, virtual image (negative image distance), height is +2.22 mm
83. Since a diverging lens has a negative focal length, by the thin lens equation, 1/di can never be positive. If di is negative, the image is on the same side as the object, so it will be virtual. The only way for di to be positive is if d0 for the object is negative. This can occur, and we will discuss HOW it can occur in class tomorrow!
AP Problems:
1989B6.
a) real
b) di = 30 cm
c) M = -0.5
d) See the image to the left.
e) focal length decreases - by Snell's law, the rays will be bent more as they enter and exit the lens, so the place where the rays intersect the principal axis will be closer to the center of the lens.
1997B5.
a) converging - the image is located behind the lens, which means image distance is positive. This can only occur with a converging lens.
b) 1/f = 1/di + 1/do so f = 22.5 mm
c) The rays will probably intersect near the 90 mm mark, but the 22.5 mm focal length makes it difficult to get it exactly. Don't worry if it isn't exactly on the 90 mm mark.
d) The image is inverted, real, and larger than the object. You can also confirm the size change by noticing that M = -di/do = -90/30 = -3.
e) The resulting image should be inverted and to the left of the mirror.
1. B
2. E
3. C
4. D
Questions:
20. If the object distance is very large, then by the thin lens equation, 1/d0 approaches zero. This means the image distance is equal to the focal length. This is where the film must be placed to form an image.
Problems:
48b. 390 mm
50. Since the image is on the other side of the lens, it must be a real image. This means that the image distance is positive. Using the thin lens equation, the focal length must be 39.8 cm.
52. 72 cm on the same side of the lens as the stamp (virtual image, negative image distance). Magnification = -4
60.
a) 15.2 cm, real image (opposite side of the lens as object), height is -2.78 mm
b) -12.1 cm, virtual image (negative image distance), height is +2.22 mm
83. Since a diverging lens has a negative focal length, by the thin lens equation, 1/di can never be positive. If di is negative, the image is on the same side as the object, so it will be virtual. The only way for di to be positive is if d0 for the object is negative. This can occur, and we will discuss HOW it can occur in class tomorrow!
AP Problems:
1989B6.
a) real
b) di = 30 cm
c) M = -0.5
d) See the image to the left.
e) focal length decreases - by Snell's law, the rays will be bent more as they enter and exit the lens, so the place where the rays intersect the principal axis will be closer to the center of the lens.
1997B5.
a) converging - the image is located behind the lens, which means image distance is positive. This can only occur with a converging lens.
b) 1/f = 1/di + 1/do so f = 22.5 mm
c) The rays will probably intersect near the 90 mm mark, but the 22.5 mm focal length makes it difficult to get it exactly. Don't worry if it isn't exactly on the 90 mm mark.
d) The image is inverted, real, and larger than the object. You can also confirm the size change by noticing that M = -di/do = -90/30 = -3.
e) The resulting image should be inverted and to the left of the mirror.
Monday, March 23, 2009
Come on Ray, just give me the equation!
A reminder: Please try the simulations below. They will go a long way to giving you an intuitive understanding of how the light rays move in both the cases of refraction and reflection. If you want a preview of tomorrow's lesson, try the lenses in the bottom simulation.
Index of Refraction simulator: http://www.ps.missouri.edu/rickspage/refract/refraction.html
Site for Simulating Mirrors (lenses too, coming soon to a 6/7 period near you!)
http://www.surendranath.org/Applets/Optics/ReflRefrCurv/CurvSurfApplet.html
Question 15 - The fish is, in reality, lower in the aquarium than is seen by the eye in the picture. Remember that the light ray from the fish is bent away from the normal when it leaves the water. This means that the actual ray from the fish to the water's surface is tilted downwards to reduce the angle from the normal to the surface.
Problems:
12. image is virtual, upright, and reduced in size. It is located at 2.09 cm behind the surface of the ball. (di = -2.09 cm).
13. magnification is -di/d0, so you can find di using the magnification of 3 (positive because it is upright.) Solving for the focal length, and multiplying by 2, the radius R = 3.9 meters.
14. concave mirror, R = 5.66 cm
27. n = c/v = 1.310
34. The light emerges from the first interface at an angle of 27.7 degrees. By geometry, we can figure out the angle at which this hits the right side of the prism. A quadrilateral is formed by the top vertex of the prism, the point where the refracted ray hits the left side, the right side intersection point, and the intersection of the normal vectors. (See the picture below)
We can see that this ray hits the other surface of the prism at 33 degrees. By Snell's Law, the ray emerges at 54.3 degrees.
35. Remember to find the angle from the normal! Theta 1 = 64.3 degrees, n1 = 1, n2 = 1.33, so theta 2 = 42.6 degrees. 2.1 meters * tan (42.6 degrees) = 1.93 meters. The total distance from the wall is therefore 2.7 m + 1.93 meters = 4.63 meters.
Index of Refraction simulator: http://www.ps.missouri.edu/rickspage/refract/refraction.html
Site for Simulating Mirrors (lenses too, coming soon to a 6/7 period near you!)
http://www.surendranath.org/Applets/Optics/ReflRefrCurv/CurvSurfApplet.html
Question 15 - The fish is, in reality, lower in the aquarium than is seen by the eye in the picture. Remember that the light ray from the fish is bent away from the normal when it leaves the water. This means that the actual ray from the fish to the water's surface is tilted downwards to reduce the angle from the normal to the surface.
Problems:
12. image is virtual, upright, and reduced in size. It is located at 2.09 cm behind the surface of the ball. (di = -2.09 cm).
13. magnification is -di/d0, so you can find di using the magnification of 3 (positive because it is upright.) Solving for the focal length, and multiplying by 2, the radius R = 3.9 meters.
14. concave mirror, R = 5.66 cm
27. n = c/v = 1.310
34. The light emerges from the first interface at an angle of 27.7 degrees. By geometry, we can figure out the angle at which this hits the right side of the prism. A quadrilateral is formed by the top vertex of the prism, the point where the refracted ray hits the left side, the right side intersection point, and the intersection of the normal vectors. (See the picture below)
We can see that this ray hits the other surface of the prism at 33 degrees. By Snell's Law, the ray emerges at 54.3 degrees.
35. Remember to find the angle from the normal! Theta 1 = 64.3 degrees, n1 = 1, n2 = 1.33, so theta 2 = 42.6 degrees. 2.1 meters * tan (42.6 degrees) = 1.93 meters. The total distance from the wall is therefore 2.7 m + 1.93 meters = 4.63 meters.
Sunday, March 22, 2009
Can YOU tell me what's on the other side of the mirror?
Hi everyone,
In case you had trouble watching the lesson in class Thursday and Friday, here is the site:
http://www.hippocampus.org/AP%20Physics%20B%20II
Click on 'Flat and Concave mirrors" under 'Geometric Optics' in the menu. You can also watch the Refraction video there as well.
Don't forget - we will have a ray tracing quiz and a problem from the handout. Feel free to check with me or each other to make sure you agree on the answers. Thanks to all of you who have already done this!
In case you had trouble watching the lesson in class Thursday and Friday, here is the site:
http://www.hippocampus.org/AP%20Physics%20B%20II
Click on 'Flat and Concave mirrors" under 'Geometric Optics' in the menu. You can also watch the Refraction video there as well.
Don't forget - we will have a ray tracing quiz and a problem from the handout. Feel free to check with me or each other to make sure you agree on the answers. Thanks to all of you who have already done this!
Wednesday, March 18, 2009
I'm Talking 'Bout the Robot In the Mirror...
Q19. Since the image is clearly magnified, this must be a concave mirror. Only concave mirrors can magnify objects. This occurs when an object is placed between the mirror and its focus.
P4. The easiest way to solve this problem is to extend the top ray through the mirror. You then obtain a pair of similar triangles. Solving the proportion, x = 75.9 cm
P11. A picture is worth a thousand words:
P4. The easiest way to solve this problem is to extend the top ray through the mirror. You then obtain a pair of similar triangles. Solving the proportion, x = 75.9 cm
P11. A picture is worth a thousand words:
P16. Leave out part b, we will discuss the equations of ray tracing tomorrow in class. Make a scale sketch of the situation on graph paper (notice it is a convex mirror!) - you should obtain that the distance of the image is between 6 and 7 centimeters on the far side of the mirror, and the image should be upright.
Tuesday, March 17, 2009
Beats, Electromagnetic Waves, and something else for dessert....
First, some web-based toys to simulate beats and other bits of interference:
http://www.mta.ca/faculty/science/physics/suren/Beats/Beats.html
See two sine waves added together.
http://micro.magnet.fsu.edu/primer/java/interference/waveinteractions2/index.html
Phase Difference Demo
http://library.thinkquest.org/19537/java/Beats.html
Simulation of beats that you can hear!
Second, the table of data you must use for calculating the speed of sound from today's experiment:
Third: Some solutions to the problems from today's homework set:
Chapter 12:
44. The unknown whistle has to be HIGHER than the 23.5 kHz if it is to be inaudible to humans. Humans can hear up to around 20 kHz. This means the whistle frequency must be 23.5 Khz + 5 kHz = 28.5 kHz.
53. 77.27 Hz is the frequency observed by the stationary tuba player - it is higher by the Doppler effect. The beat frequency would be 2.27 Hz.
54. This problem can be figured out in much the same way as 53, just with the actual frequency of the horn as the unknown. If the stationary frequency is Fo, the higher frequency perceived by the stationary observer must be (340/325)*Fo.
The equation (340/325)*Fo - Fo = 5.5 Hz will allow you to determine the actual frequency of 120 Hz. The two frequencies would be 120 Hz and 126 Hz, the 126 Hz observed from the moving car. Thus the frequency of both horns must be 120 Hz.
http://www.mta.ca/faculty/science/physics/suren/Beats/Beats.html
See two sine waves added together.
http://micro.magnet.fsu.edu/primer/java/interference/waveinteractions2/index.html
Phase Difference Demo
http://library.thinkquest.org/19537/java/Beats.html
Simulation of beats that you can hear!
Second, the table of data you must use for calculating the speed of sound from today's experiment:
Third: Some solutions to the problems from today's homework set:
Chapter 12:
44. The unknown whistle has to be HIGHER than the 23.5 kHz if it is to be inaudible to humans. Humans can hear up to around 20 kHz. This means the whistle frequency must be 23.5 Khz + 5 kHz = 28.5 kHz.
53. 77.27 Hz is the frequency observed by the stationary tuba player - it is higher by the Doppler effect. The beat frequency would be 2.27 Hz.
54. This problem can be figured out in much the same way as 53, just with the actual frequency of the horn as the unknown. If the stationary frequency is Fo, the higher frequency perceived by the stationary observer must be (340/325)*Fo.
The equation (340/325)*Fo - Fo = 5.5 Hz will allow you to determine the actual frequency of 120 Hz. The two frequencies would be 120 Hz and 126 Hz, the 126 Hz observed from the moving car. Thus the frequency of both horns must be 120 Hz.
Monday, March 16, 2009
Your interference doesn't PHASE me...
I apologize for this, but the bulk of the homework I assigned today has to do with beats, which I left out in order to talk about interference. I will be sure to give you some interference problems over the next few days to compensate.
Tomorrow's lab will hopefully take the first period, and then we can talk about beats and electromagnetic waves during 7th.
Question 17.
You can use either the equation we derived at the end of class, or reasoning about the distance traveled by each wave. If the sources are close together, then the distance traveled by each wave will be similar, and the waves will have to travel a larger difference in distance in order to reach the path difference needed for constructive or destructive interference. This means the locations will be further apart when the sources are close together.
Tomorrow's lab will hopefully take the first period, and then we can talk about beats and electromagnetic waves during 7th.
Question 17.
You can use either the equation we derived at the end of class, or reasoning about the distance traveled by each wave. If the sources are close together, then the distance traveled by each wave will be similar, and the waves will have to travel a larger difference in distance in order to reach the path difference needed for constructive or destructive interference. This means the locations will be further apart when the sources are close together.
Thursday, March 12, 2009
Introducing the Doppler 5 x 105...
Hi everyone,
Please make sure you bring in the waiver form tomorrow - if you need to download a new copy, please click here.
Solutions:
Question 17: The relative speed between the whistle and the listener is highest at point C. By today's discussion of the Doppler effect, this must also be the location of the highest frequency.
51.
a) Vobs = 370 m/s, Vsource = 340 m/s, fobs = 1959 Hz
b) Vobs = 310 m/s, Vsource = 340 m/s, fobs = 1631 Hz
52. The frequency of the wave as heard by the moving object is 46323 Hz. This frequency now acts as the source, and the bat is the observer. The new frequency heard by the bat is 43150 Hz.
78. You should be able to set up two equations with Fsource and v(the speed of the train) as the unknowns. Let Vs stand for the speed of sound.
First equation: Vs/(Vs - v) = 522/Fsource
Second equation: Vs/(Vs + v) = 486/Fsource.
Dividing the equations, Fsource drops out, and you can solve numerically to get v = 12.14 m/s.
Sites for Doppler Effect:
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=21.0
Please make sure you bring in the waiver form tomorrow - if you need to download a new copy, please click here.
Solutions:
Question 17: The relative speed between the whistle and the listener is highest at point C. By today's discussion of the Doppler effect, this must also be the location of the highest frequency.
51.
a) Vobs = 370 m/s, Vsource = 340 m/s, fobs = 1959 Hz
b) Vobs = 310 m/s, Vsource = 340 m/s, fobs = 1631 Hz
52. The frequency of the wave as heard by the moving object is 46323 Hz. This frequency now acts as the source, and the bat is the observer. The new frequency heard by the bat is 43150 Hz.
78. You should be able to set up two equations with Fsource and v(the speed of the train) as the unknowns. Let Vs stand for the speed of sound.
First equation: Vs/(Vs - v) = 522/Fsource
Second equation: Vs/(Vs + v) = 486/Fsource.
Dividing the equations, Fsource drops out, and you can solve numerically to get v = 12.14 m/s.
Sites for Doppler Effect:
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=21.0
Resonance in the Modern World
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.
And here is a video of Tuvan throat singers, as promised.
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.
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.
And here is a video of Tuvan throat singers, as promised.
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.
Wednesday, March 11, 2009
You're Standing in my Wave....
As I mentioned in class today, I expect you to be checking this blog on a daily basis to get answers for the HW assignments. This is more of an issue for those of you that are NOT staying for tutoring during 8th or after school.
below you will find a list of the last four digits of the OSIS numbers of people in the class, along with a five letter code word. Please post a comment or send me an email with your codeword to confirm you have checked the blog.
8950 - bbjah
1893 - qpkaj
3853 - zzytn
2615 - jjajs
9409 - quahl
6872 - hhsga
9021 - bnxma
6990 - jqpwo
6352 - nnzna
2531 - mqpao
1679 - mansp
9200 - gsgah
9226 - bbzna
0598 - nnqma
6706 - mpaoq
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
below you will find a list of the last four digits of the OSIS numbers of people in the class, along with a five letter code word. Please post a comment or send me an email with your codeword to confirm you have checked the blog.
8950 - bbjah
1893 - qpkaj
3853 - zzytn
2615 - jjajs
9409 - quahl
6872 - hhsga
9021 - bnxma
6990 - jqpwo
6352 - nnzna
2531 - mqpao
1679 - mansp
9200 - gsgah
9226 - bbzna
0598 - nnqma
6706 - mpaoq
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
Tuesday, March 10, 2009
Wave - You're on National Television!
For those of you on the trip today:
The homework is to read P. 322 to 328, and DO Q 10, 18, P 34, 35, 39, 41, 50
We will be having an AP problem wednesday quiz - one EMI problem, another on momentum.
Here are the quiz 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 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.
50. The string may be flat, but it is not at rest - try one of the sites below to see what happens when two reversed pulses pass each other on a string.
The homework is to read P. 322 to 328, and DO Q 10, 18, P 34, 35, 39, 41, 50
We will be having an AP problem wednesday quiz - one EMI problem, another on momentum.
Here are the quiz 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 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.
50. The string may be flat, but it is not at rest - try one of the sites below to see what happens when two reversed pulses pass each other on a string.
Monday, March 9, 2009
EMI is good for the teeth and bones.
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.
Some toys for Tuesday's lesson:
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=18 pulse superposition
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=323.0
http://www.colorado.edu/physics/2000/applets/fourier.html- Constructive/Destructive Interference
http://www.surendranath.org/Applets/Waves/TwaveRefTran/TwaveRefTranApplet.html - simple reflection
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.
Some toys for Tuesday's lesson:
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=18 pulse superposition
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=323.0
http://www.colorado.edu/physics/2000/applets/fourier.html- Constructive/Destructive Interference
http://www.surendranath.org/Applets/Waves/TwaveRefTran/TwaveRefTranApplet.html - simple reflection
Sunday, March 8, 2009
EMI solutions and a form!
Hi guys,
Thanks to all who came to robotics this weekend - it was quite a blast. I hope the REACH session was also useful for those that went. Put the next one on your calendar, March 28th at Baruch College.
Please complete the form below to let me know when you will be meeting with me.
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
Thanks to all who came to robotics this weekend - it was quite a blast. I hope the REACH session was also useful for those that went. Put the next one on your calendar, March 28th at Baruch College.
Please complete the form below to let me know when you will be meeting with me.
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
Wednesday, March 4, 2009
EMI Mr. Weinberg, I DID need to know that....
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) 17.2 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) 17.2 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
Tuesday, March 3, 2009
Field, Force, Wire, Charge - Performing This Thursday at 12:25 PM
Hmm...tomorrow is Wednesday, isn't it?
I think it's time for an AP question...think Thermo.
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. Since the force is attractive between the two wires, the current must be going in the same direction. k'I1I2/d = 6 x 10-4 N/m...solving for the unknown current gives I = 48 A
4.
a) Remember that force/unit length means that from F = IBL, force per length is F/L = IB. Thus F/L = 2.5 A * 1.10 T * sin(90) = 2.75 N/m.
b) F/L = 2.5 A * 1.10 T * sin(45) = 1.944 N/m
5. F = IBL so I = F/(BL) = 11.67 A
6. Let's define a magnetic field OUT of the page to be positive.
a) Btotal = kI1/d = (2 x 10-7 T*m/A*10 A/.05 m = 4 x 10-5T
b) Btotal = k'I1/d + k'I2/d = (2 x 10-7 T*m/A*10 A/.05 m + (2 x 10-7 T*m/A*5 A/.05 m= 6 x 10-5T
c) Btotal = k'I1/d - k'I2/d = (2 x 10-7 T*m/A*10 A/.05 m - (2 x 10-7 T*m/A*5 A/.05 m= 2 x 10-5T
7. F = IBL sin(60) = 1.212 N
8. This is an AP problem - don't be scared off by it! Remember to set a direction (in or out of the page) as positive and then sum all fields together, being sure to set fields positive or negative depending on the direction of each. We will use out of the page as positive.
Sum of fields at P:
-k'(1A)/(0.5 m) + k'(3A)/(1.0 m) = 2 x 10-7 T
This shows that the field is directed out of the page (a) at the given magnitude of 2 x 10-7 T.
c) Force is to the left, B is out of the page, and v is towards the top of the page. This only matches the left hand rule, which means the charge must be negative.
d) Fmag = qvB sin(90) so q = Fmag/(vB) = 5 x 10-7 C
e) An electric force to the right of magnitude 10-7 N would make the net force zero. Since Felec = qE, the field would have magnitude 0.2 N/C. Since the charge is negative, the electric field would need to be directed to the left for the force to be to the right.
I think it's time for an AP question...think Thermo.
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. Since the force is attractive between the two wires, the current must be going in the same direction. k'I1I2/d = 6 x 10-4 N/m...solving for the unknown current gives I = 48 A
4.
a) Remember that force/unit length means that from F = IBL, force per length is F/L = IB. Thus F/L = 2.5 A * 1.10 T * sin(90) = 2.75 N/m.
b) F/L = 2.5 A * 1.10 T * sin(45) = 1.944 N/m
5. F = IBL so I = F/(BL) = 11.67 A
6. Let's define a magnetic field OUT of the page to be positive.
a) Btotal = kI1/d = (2 x 10-7 T*m/A*10 A/.05 m = 4 x 10-5T
b) Btotal = k'I1/d + k'I2/d = (2 x 10-7 T*m/A*10 A/.05 m + (2 x 10-7 T*m/A*5 A/.05 m= 6 x 10-5T
c) Btotal = k'I1/d - k'I2/d = (2 x 10-7 T*m/A*10 A/.05 m - (2 x 10-7 T*m/A*5 A/.05 m= 2 x 10-5T
7. F = IBL sin(60) = 1.212 N
8. This is an AP problem - don't be scared off by it! Remember to set a direction (in or out of the page) as positive and then sum all fields together, being sure to set fields positive or negative depending on the direction of each. We will use out of the page as positive.
Sum of fields at P:
-k'(1A)/(0.5 m) + k'(3A)/(1.0 m) = 2 x 10-7 T
This shows that the field is directed out of the page (a) at the given magnitude of 2 x 10-7 T.
c) Force is to the left, B is out of the page, and v is towards the top of the page. This only matches the left hand rule, which means the charge must be negative.
d) Fmag = qvB sin(90) so q = Fmag/(vB) = 5 x 10-7 C
e) An electric force to the right of magnitude 10-7 N would make the net force zero. Since Felec = qE, the field would have magnitude 0.2 N/C. Since the charge is negative, the electric field would need to be directed to the left for the force to be to the right.
Sunday, March 1, 2009
Wire You Forcing Me To Stay Current?
Q11. Since the North pole is on the left, and the South pole is on the right, the B field lines go towards the right. Since the current is going into the page, and B is going to the right, by the RHR the force will be towards the top of the page.
Q17. The kinetic energy will stay constant since the magnetic force (which is also the net force acting on the particle) is always perpendicular to the velocity. This means acceleration is always perpendicular to velocity, which only occurs when an object is in uniform circular motion. Uniform circular motion is characterized by a constant speed, and by KE = 1/2mv^2, a constant kinetic energy.
17.You can find the acceleration first using kinematics, since you know displacement (1 m), final velocity (30 m/s), and initial velocity of zero since it starts at rest.
a = 450 m/s^2
Using Newton's 2nd and a FBD, you can also find that the magnetic force is the only force acting to accelerate the bar.
Fnetx = IBL = ma so I = ma/BL
I = 1.985 Amperes, magnetic field is directed into the page through the loop formed by the bars and the rails.
66.
a)
Remember that old thing called the Free Body Diagram?
We need to use it here - the magnetic force on the current accelerates the sliding wire down the rails.
Fnetx = Fmag = IBl = ma so a = IBl/m.
We can use v = vo + at to write the velocity as a function of time. The rod starts at rest, so vo = 0.
Thus v(t) = IBLt/m
b) With friction acting, the y-direction (which is directed out of the page in the photo above) becomes more important.
Fnety = Fn - mg = 0 so Fn = mg.
Since friction force Fk = mu*Fn, Fk = mu*mg
The new Fnetx = Fmag - Fk = ma
IBl - mu*mg = ma
a = (IBL/m) - mu*g
Using v = vo + at again, v(t) = 0 + ((IBL/m) - mu*g))t
c) Using the right hand rule, we can point our fingers North, and palm out of the page to go along with B. This leaves our thumb pointing to the right. This is then the direction of force.
Try it - don't guess!
Q17. The kinetic energy will stay constant since the magnetic force (which is also the net force acting on the particle) is always perpendicular to the velocity. This means acceleration is always perpendicular to velocity, which only occurs when an object is in uniform circular motion. Uniform circular motion is characterized by a constant speed, and by KE = 1/2mv^2, a constant kinetic energy.
17.You can find the acceleration first using kinematics, since you know displacement (1 m), final velocity (30 m/s), and initial velocity of zero since it starts at rest.
a = 450 m/s^2
Using Newton's 2nd and a FBD, you can also find that the magnetic force is the only force acting to accelerate the bar.
Fnetx = IBL = ma so I = ma/BL
I = 1.985 Amperes, magnetic field is directed into the page through the loop formed by the bars and the rails.
66.
a)
Remember that old thing called the Free Body Diagram?
We need to use it here - the magnetic force on the current accelerates the sliding wire down the rails.
Fnetx = Fmag = IBl = ma so a = IBl/m.
We can use v = vo + at to write the velocity as a function of time. The rod starts at rest, so vo = 0.
Thus v(t) = IBLt/m
b) With friction acting, the y-direction (which is directed out of the page in the photo above) becomes more important.
Fnety = Fn - mg = 0 so Fn = mg.
Since friction force Fk = mu*Fn, Fk = mu*mg
The new Fnetx = Fmag - Fk = ma
IBl - mu*mg = ma
a = (IBL/m) - mu*g
Using v = vo + at again, v(t) = 0 + ((IBL/m) - mu*g))t
c) Using the right hand rule, we can point our fingers North, and palm out of the page to go along with B. This leaves our thumb pointing to the right. This is then the direction of force.
Try it - don't guess!
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