Physics 1C Fall 2011: Quiz 1
Version A
1
Department of Physics
Physics 1C – Fall Quarter - 2011
Dr. Mark Paddock
INSTRUCTIONS:
1. Print your full name below
LAST NAME
FIRST NAME
MIDDLE INITIAL
2. Your code number is
THIS COURSE.
. WRITE THIS NUMBER ON EVERY SCANTRON FOR
3. Tear off bottom part of this page as indicated by the dotted line and keep for your own records. Return
the top portion with your test.
4. In compliance with the Privacy Act, your scores will be posted by code number on a website set up for
this course.
YOUR CODE NUMBER MUST BE WRITTEN ON EVERY SCANTRON FORM YOU USE FOR
THIS COURSE.
******************************************************************
Physics 1C – Fall Quarter - 2011
MY CODE NUMBER IS:
YOUR CODE NUMBER MUST BE WRITTEN ON EVERY SCANTRON FOR THIS COURSE.
TEAR OFF AT DOTTED LINE AND TURN IN BOTTOM PORTION OF THIS PAGE.
10.09.11
Physics 1C Fall 2011: Quiz 1
Version A
Department of Physics
Physics 1C Fall Quarter 2011
Dr. Paddock
Version A
DO NOT TURN OVER THIS PAGE UNTIL INSTRUCTED
TO DO SO
PUT AWAY ALL BOOKS, NOTES, PHONES, COMPUTERS
AND STUDY AIDS
INSTRUCTION – READ ME!
1. ENTER YOUR CODE NUMBER INTO THE “EXAM NUMBER”
PART OF YOUR SCANTRON.
2. ENTER THE TEST VERSION IN THE “TEST FORM” BOX ON
YOUR SCANTRON.
3. WRITE YOUR NAME LEGIBLY ON YOUR SCANTRON.
4. Flip over this page when instructed to begin exam.
5. There are 10 problems ranging in difficulty.
6. All answers should be marked on the scantron form WITH PENCIL; no
work needs to be shown.
7. An equation sheet is provided on the last page of the exam for reference.
8. Answer all questions.
GOOD LUCK!
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2
Physics 1C Fall 2011: Quiz 1
Version A
3
Closed book. No work needs to be shown for multiple-choice questions.
1. Consider a mass on a spring that is undergoing simple harmonic motion. In this simple
harmonic motion system, the magnitude of the acceleration is greatest when:
a.
b.
c.
d.
e.
the displacement is maximum.
the displacement is zero.
the speed is maximum.
the force is zero.
the speed is exactly half of its maximum value.
2. A 3.0 kg block, attached to a spring, executes simple harmonic motion according to the
position expression: x = 2.0 cos(50t), where x is in meters and t is in seconds. The spring
constant of this spring is:
a.
b.
c.
d.
e.
7,500 N/m.
1,200 N/m.
150 N/m.
100 N/m.
2,500 N/m.
3. The period of a simple pendulum on Earth is 1.00 seconds. When brought to another planet
where g is one tenth that on Earth, its period becomes:
a.
b.
c.
d.
e.
1.00 seconds.
0.316 seconds.
0.100 seconds.
10.0 seconds.
3.16 seconds.
4. A uniform cord has a mass of 0.600 kg and a length of 12.0 m. The tension in the cord is
19.6 N. What is the speed of a wave on the cord?
a.
b.
c.
d.
e.
9.90 m/s.
22.4 m/s.
4.50 m/s.
19.8 m/s.
17.3 m/s.
5. In an engine, a piston oscillates with simple harmonic motion so that its position varies
according to the expression: x = (7.5 cm) cos(4.0t + π/2), where x is in centimeters and t is in
seconds. At t = 0 seconds, the magnitude of the velocity of the piston is:
a.
b.
c.
d.
e.
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zero.
7.5 cm/s.
30 cm/s.
120 cm/s.
0.21 cm/s.
Physics 1C Fall 2011: Quiz 1
Version A
6. A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless
surface. Its total energy is 50.0 J. When the displacement is half the amplitude, the kinetic
energy is:
a.
b.
c.
d.
e.
3.13 J.
12.5 J.
25.0 J.
37.5 J.
46.9 J.
7. Consider a mass on a spring that is undergoing simple harmonic motion. As this simple
harmonic motion system goes through one complete cycle it has:
a.
b.
c.
d.
e.
constant speed.
varying amplitude.
varying period.
varying acceleration.
all of the above choices are correct.
8. A pendulum consists of a massive bob at the end of a string (as
shown to the right). If the mass of the bob is doubled, what
happens to the period of the pendulum?
a. The period of the pendulum will decrease.
b. The period of the pendulum will increase.
c. The period of the pendulum will remain the same.
9. A mass on a spring is undergoing simple harmonic motion with a frequency of 0.20 Hz. At
time t = 0 it is at the equilibrium point. At which one of the following times is it furthest in
distance from the equilibrium point?
a.
b.
c.
d.
e.
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t = 7.5 seconds.
t = 3.5 seconds.
t = 5.0 seconds.
t = 7.0 seconds.
t = 2.5 seconds.
4
Physics 1C Fall 2011: Quiz 1
Version A
5
10. A 0.20-kg block rests on a level surface and is attached to a horizontally aligned spring with
a spring constant of 40 N/m. The block is initially displaced 4.0 cm from the equilibrium
point and then released to set up a simple harmonic motion. A frictional force of 0.3 N exists
between the block and surface. What is the speed of the block when it first passes through the
equilibrium point after being released from the 4.0-cm displacement point?
a.
b.
c.
d.
e.
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0.45 m/s.
0.63 m/s.
0.80 m/s.
0.20 m/s.
0.57 m/s.
Physics 1C Fall 2011: Quiz 1
Version A
6
Equations and constants:
2
2,
(
" x = r cos ! % * r = x + y *
#
&; )
-;
-1" y %
$ y = r sin ! ' *! = tan $ ' *
# x& .
+
#
!v '
% a = !tlim
"0 !t %
%
%
$
(;
%
!x %
%v = lim %
!t "0 !t )
&
r
!F
" v = v o + at &
$
$
$!x = 12 (v o + v )t $
#
1
2' ;
$!x = v o t + 2 at $
$% v 2 = v o2 + 2 a!x $(
"0 ! f ! µ F &
s
s N$
$
$
$
# fk = µ k FN ' ;
$
$
$
% Fg = mg $
(
"
!v &
$
# !x = x f " x i ' $aavg =
!t $$
$
%%
%% $
';
$
(; #
$
% speed = d % $
!x $$
avg
v
=
%&
%
!t ) $
% avg !t $(
r
r
# F
= 0; a = 0 '%
!
%%
r
r r
r % r
r
$ ! F = ma ( ; p = mv ; P = F ! v = W ;
r
% r
%
"t
%& F2o n1 = " F1o n2 %)
k spring 2
kspring
r
m
= "kspring (#x ); a = !
A ! x 2 ); Tmass!spring = 2"
x; v =±
;
(
m
m
k
Hooke' s
I
L
1
; Tphyspend = 2!
; f = ;!=
mgd
g
T
Tpend = 2!
kspring
; x = A cos(!t + " ) = A cos(2#ft + " );
m
v=
dx
= !A" sin("t + # ) = !2$Af sin(2$ft + # );
dt
a=
d2x
2
2
2 = !A" cos("t + # ) = !(2$f ) Acos(2$ft + # ); U grav = mgh ;
dt
K = 2 mv 2 = 2 mA 2! 2 sin 2 (!t + " ); U spring = 2 kspring (!x ) = 2 k spring A 2 cos 2 ("t + # );
1
1
1
2
1
r r
1
2"
W = F !x cos" ; W nc = !E mec ; E mec = K + U grav + U spring ; E SHM = kA 2 ; kwave !
;
2
#
! = 2"f =
FT
!
2"
!
; y = A sin(kx ! "t ); v wave = !f = ; v =
; v string =
;
µ
kwave
T
T
"v
± vo %
1
v sound air at 20 o C = 343 m s ; P = 2 µ! 2 A 2v ; f ! = f $ sound
' ; (!r)con = r2 " r1 = n# ;
# v sound m v s &
#
$
(!r)des = r2 " r1 = % n +
! v $ ! n$ F
1&
() ; y = (2Asin kx )cos !t ; f n,string = nf1 = n# & = # & T ;
" 2L % " 2L % µ
2'
" v %
" v %
f n,air!open = nf1 = n$ ' n = 1,2,3,...; f n,air!closed = nf1 = n$ ' n = 1,3,5,... ; f beat = f 2 ! f1 ;
# 2L &
# 4L &
Acircle = !r ; Asphere = 4!r 2 ; g = 9.80 m/s2 ; x =
2
!b ± b 2 ! 4ac
; 1,000 m = 1 km ; 60 s = 1 min;
2a
60 min = 1 h r; 2.54 cm = 1 in ; 12 in = 1 ft ; 5, 280 ft = 1 mi ; 1,609 m = 1 mi ; 0.3048 m = 1 ft .
10.09.11