Physics 140
HOMEWORK Chapter 15 (AKA 15A)
Q4. Which of the following relationships between the acceleration a and the displacement x of a
particle involve SHM: (a) a = 0.5x, (b) a = 400 x2 , (c) a = −20 x, (d) a = −3 x2 ?
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(c). The acceleration has to be directly proportional to a negative constant times the position in SHM.
Q7. Figure 15-22 shows the x(t) curves for three experiments involving a particular spring – box system
oscillating in SHM. Rank the curves according to (a) the system’s angular frequency, (b) the spring’s
potential energy at time t = 0, (c) the box’s kinetic energy at t = 0, (d) the box’s speed at t = 0, and
(e) the box’s maximum kinetic energy, greatest first.
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(a) All equal. Same time between crests; plus, it’s the same system.
(b) 3 > 2 = 1. The Ue (0) is proportional to the square of the x value at t = 0.
(c) 1 > 2 > 3 = 0. K is proportional to the square of the slope of the x vs t curve at t = 0.
(d) Same as (c). The speed is proportional to the square root of K.
(e) 1 > 3 > 2. The maximum K will have the same value as the maximum of Ue , which corresponds
with the maximum amplitude.
P5. In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic
motion, with frequency 120 Hz. Find (a) the amplitude, (b) the maximum blade speed, and (c) the
magnitude of the maximum blade acceleration.
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Note first that ω = 2π f = 754 rad/s. Also the conversion to meters
(a) The total excursion is 2 mm, so the amplitude is half that: 1 mm = 0.001 m.
(b) v = −Aω cos ωt, so vmax = Aω = (1 mm)(754 rad/s) = 0.754 m/s.
(c) a = −Aω 2 cos ωt, so amax = Aω 2 = (1 mm)(754 rad/s)2 = 569 m/s2 .
P10. An oscillating block – spring system takes 0.75 s to begin repeating its motion. Find (a) the
period, (b) the frequency in hertz, and (c) the angular frequency in radians per second.
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(a) 0.75 s. Definition of period is time to repeat.
(b) f = 1/T = 1/(0.75 s) = 1.33 Hz. Hertz (Hz) are cycles per second.
(c) ω = 2π f = 6.28 · 1.33 = 8.38 rad/s.
P22. Figure 15-32 shows block 1 of mass 0.200 kg sliding to the right over a frictionless elevated
surface at a speed of 8.00 m/s. The block undergoes an elastic collision with stationary block 2, which
is attached to a spring of spring constant 1208.5 N/m. (Assume that the spring does not affect the
collision.) After the collision, block 2 oscillates in SHM with a period of 0.140 s, and block 1 slides off
the opposite end of the elevated surface, landing a distance d from the base of that surface after falling
height h = 4.90 m. What is the value of d?
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Recognize a multi-part problem that involves SHM, collisions, and projectile motion. To get d, we need
vf of block 1. To get vf,1 , we need the mass of block 2. To get m2 , we use SHM, needing ω to get
started.
ω = 2π/T = 44.880 rad/s.
ω 2 = k/m2 ⇒ m2 = k/ω 2 = 1208.5/(44.8802 ) = 0.60000 kg.
Use the stationary-block-2 equation for vf,1 :
vf,1 = (m1 − m2 )vi,1 /(m1 + m2 ) = (−0.4 kg)(8 m/s)/(0.8 kg) = −4 m/s.
Projectile motion: calculate the p
time to hit th p
for ∆y = −4.9 m for v0y = 0:
2
∆y = 0 − (1/2)gth ⇒ th = −2∆y/g = −2 · (−4.9 m)/(9.8 m/s2 ) = 1.000 s.
Projectile motion: vx is constant, so
d = |vx |th = (4 m/s)(1 s) = 4.00 m. Final answer to 3 sigdig.
P23. A block is on a horizontal surface (a shake table) that is moving back and forth horizontally
with simple harmonic motion of frequency 2.0 Hz. The coefficient of static friction between block and
surface is 0.50. How great can the amplitude of the SHM be if the block is not to slip along the surface?
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We need SHM and Newton’s Second Law. The force that accelerates the block is fs , just as it’s fs that
accelerates the crate in the back of the pickup truck. We will also need ω:
ω = 2π f = 6.28(2) = 12.56 rad/s.
FBD for block: mg down, n up, fs to right. The static frictional force varies sinusoidally to right and
left; I arbitrarily pick when it’s to the right. ~a to right. +x to right; +y upward.
y-eq: n − mg = 0 ⇒ n = mg.
SHM: a = −Aω 2 cos ωt, so amax = Aω 2 is the limiting case for the block.
x-eq: fs = ma. Sub for fs = fs,max = µs n = µs mg and a = Aω 2 .
µs mg = Amax ω 2 ⇒
Amax = µs g/ω 2 = (0.5)(9.8 m/s2 )/(12.56 rad/s)2 = 0.0311 m.
P30. An oscillating block – spring system has a mechanical energy of 1.00 J, an amplitude of 10.0 cm,
and a maximum speed of 1.20 m/s. Find (a) the spring constant, (b) the mass of the block, and (c) the
frequency of oscillation.
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(a) Emec = (1/2)kA2 , so k = 2Emec /A2 = 2 · (1 J)/(0.1 m)2 = 200 N/m.
2
(b) Use the fact that Kmax = Emec = (1/2)mvmax
:
2
(1/2)mvmax = 1 J ⇒ m = 2 · (1 J)/(1.2 m/s)2 = 1.39 kg.
(c) We get ω from the velocity equation and information, then use that to get f = ω/2π:
vmax = Aω = 1.20 m/s ⇒ ω = (1.20 m/s)/(0.1 m) = 12.0 rad/s.
f = 12.0/6.28 = 1.91 Hz.
P47. In Fig. 15-42, a physical pendulum consists of a uniform solid disk (of radius R = 2.35 cm)
supported in a vertical plane by a pivot located a distance d = 1.75 cm from the center of the disk. The
disk is displaced by a small angle and released. What is the period of the resulting simple harmonic
motion?
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Physical pendulum period: using Eq. 15-29. Note that
I = ICM + mh2 = (1/2)mR2 + mh2 = m(R2 /2 + h2 ).
p
p
p
T = 2π I/mgh = 2π m(R2 /2 + h2 )/mgh = 2π (R2 /2 + h2 )/gh =
p
6.28 [0.5 · (0.0235 m)2 + (0.0175 m)2 ]/[9.8 m/s2 )(0.0175 m)] = 0.366 s
P67. In Fig. 15-49, three 10 000 kg ore cars are held at rest on a mine railway using a cable that is
parallel to the rails, which are inclined at angle θ = 30◦ . The cable stretches 15 cm just before the
coupling between the two lower cars breaks, detaching the lowest car. Assuming that the cable obeys
Hooke’s law, find the (a) frequency and (b) amplitude of the resulting oscillations of the remaining two
cars.
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We know the ∆x of the cable, which acts as a spring, but need its tension to get the spring constant
k. We then use k to calculate the frequency of oscillation.
(a) FBD for the group of 3 cars just before the cable broke: mg down; n away from rails; T up-rail.
~a = 0. +x up-rail to minimize trig.
x-eq: T − mg sin θ = 0 ⇒ T = mg sin θ = (3 × 104 kg)(9.8 m/s2 )(0.5) = 1.47 × 105 N.
k = T /∆x = (1.47 × 105 N)/(0.15 m) = 9.8 × 105 N/m.
Get ω, then f .
p
p
ω = k/m = (9.8 × 105 N/m)/(2 × 104 kg) = 7 rad/s
f = ω/2π = 1.11 Hz.
(b) The equilibrium deflection from neutral with 3 cars was 15 cm, which is the initial deflection with
two cars. The equilibrium deflection with 2 cars would be 10 cm. Therefore, the initial displacement
from the new equilibrium, which is the amplitude, is 5 cm.