Physics 211 Practice Test 5
NAME:
SECTION:
Physics 211 Practice Test 5
August 31, 2011
given: I0 = 10−12 W/m2 , αsteel = 12 × 10−6 (◦ C −1 )
Multiple choice questions 2 points each.
1. When the mass of a simple pendulum is tripled, the time required for one complete
vibration
(a) increases by a factor of 3.
(b) does not change.
(c) decreases to 1/3 of its original value.
(d) decreases to 3/4 its original value.
(e) increases to 4/3 of its original value.
2. Total constructive interference occurs when two waves with similar frequency and wavelength
(a) are completely out of phase
(b) are completely in phase
(c) have 90 degrees phase difference.
(d) have a phase difference of -90 degrees.
3. An unknown tuning fork is sounded along with a tuning fork whose frequency is 256.Hz
and the beat frequency is 3.Hz. What is the frequency of the unknown tuning fork?
(a) It could be either 253Hz or 259Hz; there is no way to tell.
(b) it must be 256Hz.
(c) it must be 259Hz.
(d) It must be 253Hz
4. What is the amplitude (in meters) of a wave whose displacement is given by y =
50. sin(0.20x + 120t)cm?
(a) 0.50m
(b) 1.0m
(c) 6m
(d) 0.20m
(e) 10m
C W Fay IV: Fall2010
1
Physics 211 Practice Test 5
NAME:
5. The lower the frequency of a sound wave, the
SECTION:
(a) lower its velocity.
(b) greater its wavelength.
(c) shorter its period.
(d) greater its velocity.
6. (5 points) State the principle of superposition?
Solution: When two waves meet their amplitudes add.
7. (5 points) Give one at least one example of each of the following: (a) longitudinal wave
(b) transverse wave.
Solution: wave on a string - longitudinal ; sound wave transverse.
8. (10 points) A guitar functions by placing a wire under tension, such that its fundamental frequency is f = 196Hz if length of the wire on the guitar is l = 0.765m, and
has a density µ = 0.500g/m. The wire is attached to a peg on the head of the guitar
that is 0.500cm in diameter at 0.750cm from the base of the peg. What is the tension
in the guitar string? FT =
Solution:
f = 196Hz l = 0.765m µ = 0.500g/m d = 0.500cm
Given: 1
h = 0.750cm m = 1
v
f1 = m
2L
s
,
v=
FT
µ
(1)
s
f1
FT
m FT
=
2L µ
2Lf1 2
) = 45.0N
= µ(
m
(2)
(3)
9. (10 points) A mass-spring system is in SHM in the horizontal direction. If the mass
is 0.25kg, the spring constant is 12N/m , and the amplitude is 15cm, What would be
the speed at a half-amplitude position? v =
given: m = 0.25kg k = 12N/m A = 0.15m
s
v =
s
v =
k 2
(A − x2 ) =
m
s
k 2
A (1 − 1/4) =
m
k 2
(A − (A/2)2 )
m
s
3k
A = 0.90m/s
4m
(4)
(5)
10. (10 points) What is the ratio of power levels of two sounds with intensity levels of
40.dB and 70.dB (Assume the same distance from the source)? PB : PA =
SOLUTION:
given: β1 = 40dB β2 = 70dB
C W Fay IV: Fall2010
2
Physics 211 Practice Test 5
NAME:
SECTION:
We know that the ratio of powers is equal to the ratio intensities, and that the log of
the ratios of intensities is proportional to the difference in sound intensity levels.
P
4πR2
I1
β = 10 log( )
I0
I=
10(
β1 −β2
)
10
P1 : P2
,
P1
I1
=
I2
P2
,
β1 − β2 = 10 log(
I1
P1
=
I2
P2
= 1000 : 1
=
(6)
I1
)
I2
(7)
(8)
(9)
11. (15 points) James is listening to the same tone coming from two speakers, one 3.49m
away and the other 3.10m distant. At his position the two tones destructively interfere.
What is the lowest possible frequency if the air temperature is 20.◦ C? f =
SOLUTION:
The path length must differ by at least 1/2λ.
L1 = 3.49m L2 = 3.10m T = 20.◦ C v = 343m/s
given:
m=1
m
λ
2
v
mv
f =
=
= 439Hz
λ
2∆L
∆L =
(10)
(11)
12. (10 points) An organ pipe that is enclosed at one end has a length of 0.60m. At
20◦ C, what is the distance between a node and and adjacent anti-node for the third
harmonic? d =
SOLUTION:
given: T = 20◦ C m = 2, 3 L = 0.60m
We want to know 1/4λ
λ =
4L
m
(12)
d = 1/4λ =
l
l
= = 0.20m
m
3
(13)
13. (10 points) Suppose you want to set up a simple pendulum with a period of 1.250s
what length is required if g = 9.80m/s2 on Earth? L =
Given: T = 1.250s g = 9.80m/s2
s
T = 2π
l
g
T2
l =
g = 0.388m
4π 2
C W Fay IV: Fall2010
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(15)
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Physics 211 Practice Test 5
NAME:
SECTION:
14. (15 points) At a distance of 12.0m from a point source, the intensity level is measured
to be 70dB. At what distance from the source will the intensity level be 40dB? R =
Solution:
Given: RB = 12.0m βB = 70dB βa = 40dB
IB
IB IB
IB
) − 10 log( ) = 10 log( ∗ )
Io
Io
Io Io
IB
= 10 log
IA
IB
=
IA
βB − βA = 10 log(
(16)
βB − βA
(17)
10(
βB −βA
)
10
βB −βA
IB = IA 10( 10 )
β −β
P
P
( B10 A )
=
10
2
2
4πRB
4πRA
RA = RB 10(
βB −βA
)
20
= 380m
(18)
(19)
(20)
(21)
15. (10 points) A tuning fork vibrates at a frequency of 200Hz. If the temperature rises
from 0◦ C to 20◦ C, what is the change in the wavelength? ∆λ =
given: f = 200Hz T1 = 0◦ C T2 = 20◦ C
v2 − v1 = (331 + 0.6T2 ) − (331 + 0.6T2 ) = λ2 f − λ1 f
0.6(T2 − T1 ) = ∆λf
0.6(T2 − T1 )
= 0.060m
∆λ =
f
C W Fay IV: Fall2010
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(23)
(24)
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