Chapter 14
3.
7.
We find the spring constant from the compression caused by the increased weight:
k = mg/x = (80 kg)(9.80 m/s2)/(0.0140 m) = 5.60 × 104 N/m.
The frequency of vibration will be
f = (k/m)1/2/2π = [(5.60 × 104 N/m)/(1080 kg)]1/2/2π =
1.15 Hz.
Because the mass is released at the maximum displacement, we have
x = xmax cos ( ωt); v = – vmax sin ( ωt); a = – amax cos ( ωt).
(a) We find ωt from
v = – vmax = – vmax sin (ωt), which gives ωt = 30°.
Thus the distance is
x = xmax cos ( ωt) = xmax cos 30° =
0.866 xmax .
(b) We find ωt from
a = – amax = – amax cos (ωt), which gives ωt = 60°.
Thus the distance is
x = xmax cos ( ωt) = xmax cos 60° =
0.500 xmax .
12. (a) From the graph we see that the period is
x
0.69 s and the amplitude is 0.82 cm. We find
the spring constant from the period:
0.82 cm
T = 2π(m/k)1/2;
0.69 s = 2π[(14.3 × 10–3 kg)/k]1/2,
0.43 cm
which gives k =
1.19 N/m.
(b) For a general displacement, we have
x = A cos (ωt + φ).
We find the phase constant from the condition
at t = 0:
0.43 cm = (0.82 cm) cos φ,
0.69 s
which gives cos φ = 0.524, φ = – 58° = – 1.02 rad.
We choose the negative angle to have a positive slope at t = 0.
Thus the equation is
x = A cos [2π(t /T) + φ]
(0.82 cm) cos [(9.1 s–1)t – 1.02].
= (0.82 cm) cos {2π[t/(0.69 s)] – 1.02} =
t
19. (a) We find the frequency from
f = (k/m)1/2/2π = [(345 N/m)/(0.250 kg)]1/2/2π = 5.91 Hz, so
ω = 2πf = 2π(5.91 Hz) = 37.1 s–1.
Because the mass starts at the equilibrium position moving in the negative (downward) direction,
we have a sine function:
y = – A sin (ωt) = – (0.220 m) sin [(37.1 s–1)t].
(b) The period of the motion is
T = 1/f = 1/(5.91 Hz) = 0.169 s.
It will take one-quarter period to reach the maximum extension, so the spring will have
maximum extensions at
0.0423 s, 0.211 s, 0.381 s, … .
It will take three-quarters period to reach the minimum extension, so the spring will have
minimum extensions at
0.127 s, 0.296 s, 0.465 s, … .
20. (a) We find the frequency from the period:
f = 1/T = 1/(0.55 s) = 1.82 Hz, so
ω = 2πf = 2π(1.82 Hz) = 11.4 s–1.
The amplitude is the compression: 0.10 m. Because the mass is released at the maximum
displacement, we have a cosine function:
y = A cos (ωt) = (0.10 m) cos [(11.4 s–1)t].
(b) The time to return to the equilibrium position is one-quarter of a period:
0.14 s.
t = T = (0.55 s) =
(c) The maximum speed is
v0 = ωA = (11.4 s–1)(0.10 m) =
1.1 m/s.
(d) The maximum acceleration is
amax = ω2A = (11.4 s–1)2(0.10 m) =
13 m/s2.
The maximum magnitude of the acceleration occurs at the endpoints of the motion, so it will be
attained first at
the release point.
28. The angular frequency is
ω = 2πf = 2π(3.0 Hz) = 6.0 π s–1.
(a) The velocity at the equilibrium point is maximum:
vmax = ωA = (6.0 π s–1)(0.15 m) =
2.8 m/s.
(b) We find the velocity at the given position from
v = v0[1 – (x2/A2)]1/2
= (2.8 m/s){1 – [(0.10 m)2/(0.15 m)2]}1/2 =
2.1 m/s.
(c) We find the total energy from the maximum kinetic energy:
1.4 J.
E = Kmax = mvmax2 = (0.35 kg)(2.8 m/s)2 =
(d) Because the mass starts at the maximum position, we use a cosine function:
x = A cos ωt = (0.15 m) cos (6.0 π s–1)t.
32. (a) We find the spring constant from
2πf = (k/m)1/2;
2π(3.5 Hz) = [k/(0.240 kg)]1/2, which gives k =
1.2 × 102 N/m.
(b) We find the total energy from the maximum potential energy:
0.12 J.
E = Umax = kA2 = (1.16 × 102 N/m)(0.045 m)2 =
35. For the collision of the bullet and block momentum is conserved:
mv = (m + M)V, so V = mv/(m + M).
The kinetic energy of the bullet and block immediately after the collision is stored in the potential energy of the
spring when the spring is fully compressed:
(m + M)V2 = kA2;
(m + M)[mv/(m + M)]2 = kA2;
[(0.007870 kg)v]2/(0.007870 kg + 6.023 kg) = (142.7 N/m)(0.09460 m)2, which gives v =
352.6 m/s.
43. We assume that 14° is small enough that we can consider this a simple pendulum, with a period
T = 2π(L/g)1/2 = 2π[(0.30 m)/(9.80 m/s2)]1/2 = 1.10 s, and ω = 2π/T = 2π/(1.10 s) = 5.71 s–1.
Because the pendulum is released at the maximum angle, the angle will oscillate as a cosine function:
θ = θ0 cos (ωt) = (14°) cos [(5.71 s–1)t].
(a) θ = (14°) cos [(5.71 s–1)(0.65 s)] =
– 12°.
This is reasonable, because the time is slightly over half a period.
(b) θ = (14°) cos [(5.71 s–1)(1.95 s)] =
+ 1.9°.
(c) θ = (14°) cos [(5.71 s–1)(5.00 s)] =
– 13°.
This is reasonable; the time is 4 periods and 0.60 s, so it should be close to the answer for part (a).
44. We use energy conservation between the release point and the
lowest point:
Ki + U i = Kf + U f ;
0 + mgh = mv02 + 0, or v02 = 2gh = 2gL(1 – cos θ).
When we use a trigonometric identity, we get
v02 = 2gL(2 sin2 θ).
θ
L cos θ
For a simple pendulum θ is small, so we have sin θ ≈ θ.
Thus we get
v02 = 2gL2 ( θ)2, or
v0 = θ(gL)1/2.
h
L
69. (a) For the frequency we have
f = (g/L)1/2(1/2π) = [(9.80 m/s2)/(0.63 m)]1/2(1/2π) =
0.63 Hz.
(b) We use energy conservation between the release point and the
lowest point:
Ki + U i = Kf + U f ;
0 + mgh = mv02 + 0;
(9.80 m/s2)(0.63 m)(1 – cos 15°) = v02,
0.65 m/s.
which gives v0 =
(c) The energy stored in the oscillation is the initial potential energy:
Ui = mgh = (0.365 kg)(9.80 m/s2)(0.63 m)(1 – cos 15°) =
0.077 J.
Note that this is also the maximum kinetic energy, mv02.
θ0
L cos θ0
h
L
78. (a) We find the speed from energy conservation:
Ki + U i = Kf + U f ;
Mv02 + 0 = 0 + kA2;
(5.0 kg)v02 = (310 N/m)(0.24 m)2, which gives v0 =
1.9 m/s.
(b) We find the period of the oscillation from
T = 2π(m/k)1/2 = 2π[(5.0 kg)/(310 N/m)]1/2 = 0.798 s.
The car will be in contact with the spring for half a cycle, so the time is
t = T = (0.798 s) = 0.40 s.
88. When a mass m is a distance x from the center of the Earth,
there will be a gravitational force only from the mass
within a sphere of radius x. This mass is
M′ = (ME/ πrE3) πx3 = MEx3/rE3.
The force will be toward the center, opposite to x:
F = – GM′m/x2 = – G(MEx3/rE3)m/x2 = – (GMEm/rE3)x.
Thus we see that the restoring force is proportional to the
displacement, so the motion will be simple harmonic, with
keff = GMEm/rE3.
The apple will take one period to return to the initial location:
T = 2π(m/keff)1/2 = 2π[m/(GMEm/rE3)]1/2 = 2π(rE3/GME)1/2
= 2π[(6.38 × 106 m)3/(6.67 × 10–11 N · m2/kg2)(5.98 × 1024 kg)]1/2
= 5.07 × 103 s =
84.5 min.
m
x
rE
Chapter 15
1.
The speed of the wave is
v = fλ = λ/T = (9.0 m)/(4.0 s) =
2.3 m/s.
7.
We find the tension from the speed of the wave:
v = [FT/(m/L)]1/2;
(4.8 m)/(0.85 s) = {FT/[(0.40 kg)/(4.8 m)]}1/2 , which gives FT =
2.7 N.
9.
(a) Because the pulse travels up and back, the speed is
v = 2L/t = 2(600 m)/(16 s) =
75 m/s.
(b) The mass density of the cable is
µ = m/L = ρAL/L = ρA.
We find the tension from
v = (FT/µ)1/2 = (FT/ρA)1/2;
75 m/s = [FT/(7.8 × 103 kg/m3)π(0.75 × 10–2 m)2]1/2 , which gives FT =
7.8 × 103 N.
12. Because the speed, frequency, and medium are the same for the two waves, the intensity depends on the
amplitude only:
I ∝ D M 2.
For the ratio of intensities we have
I2/I1 = (DM2/DM1)2 ;
2 = (DM2/DM1)2 , which gives DM2/DM1 =
1.41.
14. We assume that the wave spreads out uniformly in all directions.
(a) The intensity will decrease as 1/r2, so the ratio of intensities is
I2/I1 = (r1/r2)2;
I2/(2.2 × 106 W/m2) = [(100 km)/(4.0 km)]2 , which gives I2 =
1.4 × 109 W/m2.
(b) We can take the intensity to be constant over the small area, so we have
P2 = I2S = (1.38 × 109 W/m2)(5.0 m2) =
6.9 × 109 W.
20. The traveling wave is
D = (0.48 m) sin [(5.6 m–1)x + (84 s–1)t].
(a) We find the wavelength from the coefficient of x:
(5.6 m–1)x = 2πx/λ, which gives λ =
1.12 m.
(b) We find the frequency from the coefficient of t:
13 Hz.
(84 s–1)t = 2πft, which gives f =
(c) From the positive sign between the x and t terms, the wave is traveling in the – x direction, with
speed
v = fλ = (13.4 Hz)(1.12 m) =
15 m/s (toward negative x).
(d) The amplitude is the coefficient of the sine function:
DM =
0.48 m.
(e) The speed of a particle is
u = ∂D/∂t = (0.48 m)(84 s–1) cos [(5.6 m–1)x + (84 s–1)t] = (40 m/s) cos [(5.6 m–1)x + (84 s–1)t].
Thus we have
umax = (40 m/s)(1) =
40 m/s;
umin = (40 m/s)(0) =
0.
32. (a) For the traveling wave in the lighter cord,
D1 = (0.050 m) sin [(6.0 m–1)x – (12.0 s–1)t],
we find the wavelength from the coefficient of x:
1.05 m.
(6.0 m–1)x = 2πx/λ1 , which gives λ1 =
(b) We find the tension from the velocity:
v1 = ω/k1 = (FT/µ1)1/2;
(12.0 s–1)/(6.0 m–1) = [FT/(0.10 kg/m)]1/2, which gives FT =
(c) The tension and frequency do not change, so we have
v2 = ω/k2 = (FT/µ2)1/2, or
k1/k2 = λ2/λ1 = (µ1/µ2)1/2;
λ2/(1.05 m) = [(0.10 kg/m)/(0.20 kg/m)]1/2, which gives λ2 =
0.40 N.
0.74 m.
36. From the diagram the initial wavelength is 2L, and the
final wavelength is 3L/2. The tension has not changed,
so the velocity has not changed:
v = f1λ1 = f2λ2;
(294 Hz)(2L) = f2(3L/2), which gives f2 =
392 Hz.
L
Unfingered
Fingered
42. We find the speed of the wave from
v = [FT/(m/L0)]1/2 = {(520 N)/[(0.0036 kg)/(0.900 m)]}1/2 = 361 m/s.
The wavelength of the fundamental for a string is λ1 = 2L. We find the fundamental frequency from
f1 = v/λ1 = (361 m/s)/2(0.60 m) =
300 Hz.
All harmonics are present so the first overtone is the second harmonic:
f2 = (2)f1 = (2)300 Hz =
600 Hz.
The second overtone is the third harmonic:
900 Hz.
f3 = (3)f1 = (3)300 Hz =
44. The hanging weight creates the tension in the string: FT = mg. The speed of the wave depends on the tension
and the mass density:
v = (FT/µ)1/2 = (mg/µ)1/2.
The frequency is fixed by the vibrator, so the wavelength is
λ = v/f = (1/f )(mg/µ)1/2. With a node at each end, each loop corresponds to λ/2.
(a) For one loop, we have λ1/2 = L, or
2L = v1/f = (1/f )(m1g/µ)1/2;
2(1.40 m) = (1/60 Hz)[m1(9.80 m/s2)/(4.8 × 10–4 kg/m)]1/2, which gives m1 =
1.4 kg.
(b) For two loops, we have λ2/2 = L/2, or
L = v2/f = (1/f )(m2g/µ)1/2;
1.40 m = (1/60 Hz)[m2(9.80 m/s2)/(4.8 × 10–4 kg/m)]1/2, which gives m2 =
0.35 kg.
(c) For five loops, we have λ5/2 = L/5, or
2L/5 = v5/f = (1/f )(m5g/µ)1/2;
2(1.40 m)/5 = (1/60 Hz)[m5(9.80 m/s2)/(4.8 × 10–4 kg/m)]1/2, which gives m5 =
0.055 kg.
The amplitude of the standing wave can be much greater than the vibrator amplitude because of the
resonance built
63. An object will leave the ground when the maximum acceleration of the ground during the SHM as the wave
passes is greater than the acceleration due to gravity:
amax = DM ω2 > g, or
DM > g/ω2 = g/4π2f 2 = (9.80 m/s2)/4π2(0.50 Hz)2 =
0.99 m.