This is a practice test for Test #2 in Physics 202. Problems are in italics, hints and a couple of more detailed solutions are in bold, solution attempts by students in regular type.
Suggested use: Work through the practice test without looking at hints/solutions, then look at hints/solutions and self-critique in the usual manner, inserting your responses into the document. Be sure to mark insertions before and after with **** so your instructor can find them quickly.
Notes for specific classes:
course Phy 202
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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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The given student solution is correct.
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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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dB = 10 log(I / I_0), so
10^(dB / 10) = I / I_0 and
I = I_0 * 10^(dB / 10). Thus, as in the student solution,
I = 1.0 x 10^-12 w/m^2 x (10^4.6)
= 4.0 x 10^-8 w / m^2
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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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<h3>This is reasoned out in the same way as #5.</h3>
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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
<h3>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.</h3>
Problem Number 9
When held under tension about 170 Newtons a chain with total mass about 15 kg and length 8 meters supported a pulse which made 3 round trips in 5.4 seconds.
Based on the time and distance traveled by the pulse:
Round trip is 16 m, total for 3 round trips is 48 m.
48 m / (5.4 sec) = 9 m/s ballpark.
Based on tension and mass/unit length:
c = sqrt( 170 N / (15 kg / 8 m) )
= sqrt( 170 kg m/s^2 / (1.88 kg / m) )
= sqrt(90 m^2 / s^2)
= 9.5 m/s.
University Physics Problem: Show that the function y(x, t) = .64 e^-( 550 t - .58 x)^2 satisfies the wave equation, and give the frequency, wavelength and velocity of the wave. Sketch the wave from x = - 15 to x = 15 at t=0 and at t = .008463.