#$&*
course Phy 202
Please have back by 6 pm on 7/21!! I will take this exam thursday morning
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Problem Number 1
A string is under a tension of 13 Newtons and lies along the x axis. ""Beads"" with mass 4.2 grams are located at a spacing of 18 cm along a light but strong string. At t = 0 bead A is moving in the y direction at .128 m/s, and this bead is at at y position .0017 meters, while the bead to its right is at y position .0014 meters and the bead to its left at y position .0012 meters. Find:
• the acceleration of the given bead
• its approximate velocity .031 seconds later and
• the distance it will move in the .031 seconds.
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.I have to say, I honestly have NO idea what equation to even start with this one. Also, since the two chapters we are on have to do with light and sound, how does this question even fit in?
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This problem should not occur on a Phy 202 test, and would not be counted against you. The problem has everything to do with wave motion, but it's beyond the scope of the 202 course.
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Problem Number 2
If a traveling wave has wavelength 2.9 meters and the period of a cycle of the wave is .051 second, what is the propagation velocity of the wave? What is its frequency?
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.Frequency = speed/wavelength
= 343/2.9
= 118 Hz.
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Problem Number 3
If hearing threshold intensity is 10^-12 watts/m^2, then what is the intensity of a sound which measures 46 decibels?
. I= I0 x 10 ^(4.6)
I= 1.0 x 10^-12 w/m^2 x (10^4.6)
= 4.0 x 10^-8
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Problem Number 4
Two sources separated by 2.45 meters emit waves with wavelength .43 meters emit waves in phase. The waves travel at identical velocities to a distant observer. At any point along the perpendicular bisector of the line segment connecting the two points, the two waves will arrive in phase an hence reinforce. At what nonzero angle with the perpendicular bisector will the first interference maximum be observed?
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.frequency= speed/wavelengh
= 343/ 0.43
= 798 Hz
I’m not really sure where I’m supposed to go, or if I even need the frequency
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If the waves make angle theta with the perpendicular bisector, then one wave travels distance a sin(theta) more than the other, where a is the separation of the two points. This is called the path difference.
Interference maxima occur when the path difference is a whole number of wavelengths, in which case the peaks of one wave meet the peaks of the other and they reinforce.
Interference minima occur when the path difference is a whole number of wavelengths, plus half a wavelength, in which case the peaks of one meet the valleys of the other and they cancel.
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Problem Number 5
A string of length 3 meters is fixed at both ends. It oscillates in its fundamental mode with a frequency of 138 Hz and amplitude .9 cm. If its mass is 3 grams, then what tension is it under?
.F=ma
F= 3 g (138x 0.09)
= 37.26
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The velocity of the disturbance is v = sqrt( T / (m/L) ), where T is tension, and m/L is mass per unit length.
The fundamental mode of a string fixed at both ends has node-antinode configuration N A N, so the length of the string is two quarter-wavelengths or 1/2 wavelength. The string is 3 meters long, so the wavelength is 6 meters.
You therefore know the frequency and wavelength of the fundamental. From this you can find the velocity of propagation.
Knowing the mass m, the length L and the velocity v you can solve v = sqrt( T / (m/L) ) for the tension T.
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Problem Number 6
A sound source with frequency 475 Hz moves away from an observer at 12 m/s. If the speed of sound is 340 m/s, then what frequency will be heard by the obsever?
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.475 Hz/ 12
= 39.6 Hz
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This situation is described by the Doppler shift, either by the reasoning used in the Introductory Problem Set or using the equations given there and in the text.
v_sound is 340 m/s
v_source is -12 m/s
f_source, or just f, is 475 Hz.
f_observed, or f ', is found using the equation.
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Problem Number 7
The fundamental harmonic in a uniform string of length 7 meters, having mass 30 grams, is 43 Hz. What is the tension in the string?
.30 x 7 x 43 ?? I’m really not sure of the correct equation to use..
=9030
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This is reasoned out in the same way as #5.
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Problem Number 8
A person 1.38 meters high stands 3 meters from a converging lens. A real inverted image forms 2.8 meters on the other side of the lens.
• What is the focal length of the lens?
• What is the size of the image?
• Sketch a diagram explaining how the image forms.
1. f= r/2
= 1.5/2
= 0.75
2. ?? I’m not sure of an equation to use here. Maybe take the height of the person and divide by the focal length?
1.38/0.75
=1.84 m
1/f = 1/i + 1/o. The image distance is i = 2.8m, the object distance is o = 3 meters, so you can find the focal length f.
The size of the image is in the same proportion to the size of the object and the image distance to the object distance.
f = r/ 2 applies to a spherical mirror with radius of curvature r. That doesn't help here, where the optical device is a lens rather that a mirror.