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Answers
11.1 Simple
Harmonic Motion
Practice A, Hook’s Law, pg 371
[1a] 15 N/m [b] less stiff [2] 320 N/m [3] 2700 N/m [4] 81 N
Ch 11 Review Q, pg 396:
1] oscillation about an equilibrium position in which a restoring force is
proportional to displacement.
2] mass-spring system, child on a swing, pendulum of a grandfather clock,
metronome.
3] No, acceleration changes throughout the oscillator’s ,option. It is zero at
the equilibrium position and greatest at maximum displacement.
4] No, a pendulum’s displacement is approximately proportional to its restoring
force only at angles smaller than 15o.
5] gravitational potential energy. When April lets the bob go,
PE = max & KE = 0. At the bottom of its swing, KE = max and PE = 0.
6] because frictional forces are neglected in an ideal mass-spring system.
7] the tangent component; because it always pulls the bob toward the equilibrium
position.
8] 130 N/m 9] 580 N/m
Section 1.1 Review Q, pg 375:
[1] c [2] 0.52 N
[3] Force & acceleration decrease, velocity increases
[4] because the acrobat’s momentum carries her through the
equilibrium position.
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11.2 Measuring Simple Harmonic Motion
Practice B, SHM of a Pendulum, pg 379
[1] 140 m [2] 25 cm [3] 3.6 m
[4a] 3.749 s, 0.2667 Hz [b] 3.754 s, 0.2664 Hz
[c] 3.758 s, 0.2661 Hz
Practice C, SHM of a Mass-Spring System, pg 381
[1] 210 N/m [2] 25 N/m [3] 39.7 N/m [4] 0.869 s
[5a] 1.7 s, 0.59 Hz [b] 0.14 s, 7.21 Hz [c] 1.6 s, 0.62 Hz
Ch 11 Review Q, pg 397: 19] 9.7 m
[20a] 2.000 s [b] 9.812 m/s2 [c] 9.798 m/s2
[21a] 0.57 s [b] 1.8 Hz
Section 11.2 Review Q, pg 381:
[1] 3.0 Hz, 0.33 s [2a] 3.2 s [b] 0.31 Hz
[3a] 25 N/m [b] 1.1 s
[4] The system with the larger mass has the greater period.
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11.3 Properties of Waves
Practice D, Wave Speed, pg 387
[1] 0.081 m < λ < 12 m
[2a] 3.41 m [b] 5.0x10-7 m [c] 10x10-10
m
[3] 4.74x1014 Hz [4a] 346 m/s [b] 5.86 m
Ch 11 Review Q, pg 397: 35] 0.0333 m
Section 11.3 Review Q, pg 388:
[1] The disturbance moves, not the medium.
[2a] One portion of the spring should have a single compressed region
and a single stretched region.
[b] The spring should have several compressed regions and several
stretched regions.
[c] The spring should contain a single hump either above or below
its equilibrium position.
[d] The spring should contain several humps above and below its
equilibrium position.
[3] The graph for (b) should look like Fig 11(b) but should have a
y-axis labeled density. Graph for (d) should resemble Fig 11(b)
[4] Energy will be 16 x as great [5] 60,000 Hz
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11.4 Wave
Interactions
Ch 11 Review Q, pg 398:
40] Yes, because waves do not collide like matter; they add to form a
resultant wave.
41] Zero
42] Yes; when constructive interference occurs.
43] a, b, & d (λ = 0.5 L, L and 2L respectively)
Section 11.4 Review Q, pg 394:
[1] 0.50 m [2] 0.00 m, destructive [3] 0.30 constructive
[4] Wavelengths that will produce standing waves include 4.0 m,
2.0 m, and 1.3 m; Any value that does not allow both ends of
the string to be node is acceptable.
[5] 4 nodes; 3 antinodes.
Ch 11 Mixed Review
[44] 14 N [45] 1.7 N [46] 2.0 Hz, 0.50 s, 0.30 m/s
[47] 446 m [48] 0.129 m < λ < 1.73 m [49] 9.70 m/s2
[50] 5.17x1014 Hz [51]
9:48 A.M.
Standardized Test Prep, pg 400
1] C [2] J [3] C [4] F [5] C [6] G [7] B [8] G [9] A
[10] H [11] A [12] H [13] A [14] J [15] A [16] G
[17] 5.77X1014 Hz, 1.73x10-15
s [18] electromagnetic waves
[19] Possible correct answers include 4.0 m, 2.0 m, 1.3 m, 1.0 m, or other
wavelengths such that nλ = 4 m (where n is a positive integer).
[20] 22.4 m [21] 0.319 m
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1 How much force is required to pull a spring 3.0 cm
from its equilibrium position if the spring constant is 2700 N/m?
2 A mass of 0.55 kg hung on a spring stretches 36 cm
form its equilibrium position.
a] Draw & label the diagram.
b] Find the spring constant.
3 How far (cm) will a 10 N force compress a spring with
a constant of
2.5 N/m?
4 A 2.0 m pendulum suspends a 3.0 kg mass.
a] Draw and Label the vector diagram when the pendulum is at 10o with
the vertical.
b] Find the restoring force when it is at 10o with the vertical.
c] Draw and Label the vector diagram when the pendulum is at 75o with
the horizontal.
d] Find the restoring force when it is at 75o with the horizontal.
5 If the period of the pendulum is 24 s, how long is it?
6 Calculate the period and frequency of a 3.500 m
pendulum at the north pole where g = 9.832 m/s2.
7 A pendulum moves through its equilibrium position once
every 1.000 s.
a] What is the period of this pendulum?
b] At location X this pendulum is 0.9942 m long. What is the free-fall
acceleration at location X?
8 A 125 N object makes 20 complete vibrations is 4.0 s
when hanging from a spring.
a] Find the period.
b] What is the spring constant?
9 A spring of spring constant 30.0 N/m is attached to a
2.30 kg object, and the system is set in motion.
Find the period and frequency of vibration.
10 The red light emitted by a He-Ne laser has a
wavelength of 633 nm in air and travels at 3.00x108
m/s.
Find the period and frequency of the laser light.
11 A tuning fork produces a sound with a frequency of 256
Hz and a wavelength in air of 1.35 m
a] What value does this give for the speed of sound in air?
b] What would be the wavelength of this same sound in water in which sound
travels at 1500 m/s?
Ans: 1] 81 N 7a] 2 s [b] 9.812
m/s2
2] 15 N/m 8] 0.20 s/cycle, 13000 N/m
3] 400 cm 9] 1.74 s/cycle, 0.575 cycles/s
4b] 5.1 N [c] 7.6 N 10] 4.73x1014
cycles/s,
5] 143 m 0.211x10-14
s/cycles
6] 3.749 s/cycle, 0.2668 cycles/s 11] 345.6 m/s, 5.86 m
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1 How much force is required to pull a spring 3.0 cm
from its equilibrium position if the spring constant is 2700 N/m?
G/F: F = ?, X = 3 cm x 1 m/100 cm = 0.03 m, k = 2700 N/m
F = -kX
F = - 2700 x 0.03 = 81 N opposite the stretch direction.
2 A mass of 0.55 kg hung on a spring stretches 36 cm
form its equilibrium position.
a] Draw & label the diagram.
b] Find the spring constant.
G/F: m = 0.55 kg, X = 36 cm x 1 m/100 cm = 0.36 m, k = ?
Hung-->F = mg = 0.55 x 9.81 = 5.4 N down
F = - kX--->
-5.4 N = -k 0.36 m--->k = 5.4/0.36 = 15 NN/m
3 How far (cm) will a 10 N force compress a spring with
a constant of
2.5 N/m?
G/F: X = ?; F = 10 N, k = 2.5 N/m
F = -kX
10 N = -2.5X--->X = 10 N/2.5 N/s = 4 m x 100 cm/1 m = 400 cm
4 A 2.0 m pendulum suspends a 3.0 kg mass.
a] Draw and Label the vector diagram when the pendulum is at 10o with
the vertical.
b] Find the restoring force when it is at 10o with the vertical.
Restoring force
= the component of the weight tangent to the
motion (see diagram). Weight: W = mg = 3x9.81 = 29.43 N
F = W Sin(θ) = 3x9.81*Sin(10) = 5.1 N
c] Draw and Label the vector diagram when the pendulum is at 75o with
the horizontal.
d] Find the restoring force when it is at 75o with the horizontal.
F = W Cos(θ) = 3x9.81*Cos(75) = 7.6 N
5 If the period of the pendulum is 24 s, how long
is it?
T = 2π√L/g
24 = 2π√L/9.81--->Square
both sides--->576 = 39.5 L/9.81
L = 576x9.81/39.5 = 143 m
6 Calculate the period and frequency of a 3.500 m
pendulum at the north pole where g = 9.832 m/s2.
T = 2π√L/g =
6.283√3.500/9.832 = 3.749 s/cycle
f = 1/T = 0.2668 cycles/s
7 A pendulum moves through its equilibrium position once
every 1.000 s.
a] What is the period of this pendulum?
b] At location X this pendulum is 0.9942 m long. What is the free-fall
acceleration at location X?
T = 2π√L/g--->2
= 2π√0.9942/g
--->Square both sides--->4 = 39.48(0.99942/g)
--->g = 39.48(0.9942/4) = 9.812 m/s22
8 A 125 N object makes 20 complete vibrations is 4.0 s
when hanging from a spring.
a] Find the period.
G/F: W = 125 N = mg---m = 125/9.81 = 12.7 kg
f = 20 Cycles/4 s = 5; T = ?
T = 1/f = 1/5 = 0.2 s/cycle
b] What is the spring constant?
T = 2π√m/k--->0.2
= 6.283√12.7/k--->Square both sides--->
0.04 = 39.48(12.7/k)--->k = 39.48(12.7)/0.04 = 12535 N/m
9 A spring of spring constant 30.0 N/m is attached to a
2.3 kg object, and the system is set in motion.
Find the period and frequency of vibration.
T = 2π√m/k =
6.283√2.3/30 = 1.74 s/cycle
f = 1/T = 1/1.74 = 0.575 cycles/s
10 The red light emitted by a He-Ne laser has a
wavelength of 633 nm in air and travels at 3.00x108
m/s.
Find the period and frequency of the laser light.
G/F: λ = 633 nm = 633x10-9
m, V = 3.00x108 m/s, f = ?
V = λf--->3.00x108 = 633x10-9
f
--->f = 3.00x108/633x10-9
= 0.00473x1017 = 4.73x1014
cycles/sec
T = 1/f = 1/4.73x1014
cycles/sec = 0.211x10-14
s/cycles
11 A tuning fork produces a sound with a frequency of 256
Hz and a wavelength in air of 1.35 m
a] What value does this give for the speed of sound in air?
V = λf = 1.35x256 = 345.6 m/s
b] What would be the wavelength of this same sound in water in which sound
travels at 1500 m/s?
V = λf--->1500 = λx256--->λ = 1500/256 = 5.86 m
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