# In Fig rail moment of inertia

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```					                                    8.   Oscillations

Chapter 8            Oscillations

Basic Requirements

1. Understand the concept of Simple Harmonic Motion (SHM);
2. Master the force law for Simple harmonic motion;
3. Understand the meaning of phase, in phase and out of phase of SHM;
4. Master the characteristic of composition of simple harmonic motion in the same
direction with the same frequency;
5. Understand the superposition of two SHMs in the directions perpendicular to each
other;
6. Understand damped simple harmonic motion;
7. Understand forced oscillations and resonance.

Review and Summary

Frequency The frequency f of periodic or oscillatory motion is the number of
oscillations per second. In the SI system, it is measured in hertz:

1 1hertz= 1Hz = 1 oscillation per second = 1s 1                     (8-1)

Period The period T is the time required for one complete oscillation, or cycle. It
is related to the frequency by

1
T                                            (8-2)
f

Simple Harmonic Motion        In simple harmonic motion (SHM), the displacement
x(t ) of a particle from its equilibrium position is described by the equation
x  xm cos(t   ) (displacement)                   (8-3)

in which xm is the amplitude of the displacement, the quantity (t   ) is the phase
of the motion , and  is the phase constant. The angular frequency  is related to
the motion by
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8.   Oscillations
2
       2 f      (angular frequency)                   (8-5)
T
Differentiating Eq. 8-3 leads to equations for the particle’s velocity and acceleration
during SHM as function of time:

v   xm sin(t   )                                  (8-6)

and     a   2 xm cos(t   )                                (8-7)

In Eq. 8-6, the positive quantity  xm is the velocity amplitude vm of the motion. In
Eq. 8-7, the positive quantity  xm is the acceleration amplitude am of the
2

motion.

The Linear Oscillation      A particle with mass m that moves under the influence of a
Hooker’s law restoring force given by F  kx exhibits simple harmonic motion
with

k
        (angular frequency)                          (8-12)
m

m
and T  2        (period)                                  (8-13)
k
Such a system is called a linear simple harmonic motion oscillator.

Energy    A particle in simple harmonic motion has, at any time, kinetic energy
1 2                               1
K    mv and potential energy U  kx 2 . If no friction is present, the
2                                 2
mechanical energy E  K  U remains constant even though K and U change.

Pendulums Examples of devices that undergo simple harmonic motion are the
torsion pendulum of Fig. 16-7, the simple pendulum of Fig. 8-9, and the physical
pendulum of Fig. 8-10. Their periods of oscillation for small oscillations are,
respectively,

T  2 I                                         (8-22)

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8.   Oscillations

T  2 I mgh                                          (8-26)

and     T  2 L g                                             (8-27)

Simple Harmonic Motion and Uniform Circular Motion Simple harmonic motion
is the projection of uniform circular motion onto the diameter of the circle in which
the latter motion occurs. Figure 8-14 shows that all parameters of circular motion
(position, velocity, and acceleration) project to the corresponding values for simple
harmonic motion.

Damped Harmonic Motion            The mechanical energy E in a real oscillating system
decreases during the oscillations because external forces, such as a drag force, inhibit
the oscillations and transfer mechanical energy to thermal energy. The real oscillator
and its motion are then said to be damped. If the damping force is given by
Fd  bv , where v is the velocity of the oscillator and b is a damping constant, then
the displacement of the oscillator is given by

x(t )  xm e  bt 2 m cos( 't   )                         (8-40)

where    ' , the angular frequency of the damped oscillator, is given by
k  b2
'           2                                     (8-41)
m 4m

If the damping constant is small ( b            km ), then  '   , where  is the angular
frequency of the undamped oscillator. For small b, the mechanical energy E of the
oscillator is given by
1 2 bt m
E (t )      kxme                                     (8-42)
2

Force Oscillations and Resonance                 If an external driving force with angular
frequency  d acts   on the oscillating system with natural angular frequency  , the
system oscillates with angular frequency  d . The velocity amplitude vm of the
system is greatest when

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8.   Oscillations

d                                             (8-44)

a condition called resonance. The amplitude xm of the system is (approximately)
greatest under the same condition.

Examples

Example 1     Suppose that two springs in figure have different spring constants k1 and
k2. Show that the frequency f of oscillation of the block is then given by
f     f12  f 22 where f1 and    f2 are the frequencies at which the block would
oscillate if connected only to spring 1or only to spring 2.
Solution：The key idea is that the motion of block-spring
system is SHM. Assuming that the spring 1 is stretched               k1           k2
for a small displacement x and the spring 2 is compressed                   m
by the same amount at the same time. From the Hook’s
law we can write                                                  Fig.8-1 Example 1.

Fx  k1x  k 2 x=  (k1  k 2 )x

where (k1  k 2 ) is the coefficient of a composite spring. Using the Newton’s Second

Law, we can obtain

d2x
m         (k1  k 2 )x
dt 2

d2x
m         (k1  k 2 )x=0
dt 2
Thus the angular frequency is

k1  k 2

m
And the frequency f of oscillation of the oscillator is

   1        k1  k2    1
f                            12  2 
2
2 2           m      2

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8.   Oscillations

Example 2 A block-spring system of the constant k  9.8N/ m , and the block of mass
m  20 g . Now the spring is stretched 2 2cm away from the equilibrium position,
and the block has the initial speed the 7.0m / s . Find the kinematics equation of the
oscillator. (SI)
We
Solution： choose the equilibrium position as the origin and the x axis horizontal to
the right. So the equation of the spring oscillator is

x  A cos(0t   )                                     (1)

k  9.8( N / m), m  200 g
k      9.8                                      (2)
0                  7(rad / s)
m   200 103
From the avaliable conditions given by the problem, we know

t  0  x  x0  2 2(cm), v  v0  7.0(cm / s)
so we can obtain

 02
A  x  2  3 102 (m)
2
(3)
0
0

x0  A cos      
    0.34(rad )                                (4)
v0   A0 sin  
Substituting the Eq.(2)-(4) into the Eq.(1), the kinematics equation of the oscillator can
be written as:

x  3 102 cos(7t  0.34)                      (Answer)

Example 3 An object oscillates with simple harmonic motion along the x axis. Its
displacement from the origin varies with time according to the equation

x  (4.00m) cos( t  )
4
where t is in seconds and the angles in the parentheses are in radians. (a) Determine
the amplitude, frequency, and period of the motion. (b) Calculate the velocity and
acceleration of the object at any time t. (c) Using the results of part (b), determine the
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8.   Oscillations
position, velocity, and acceleration of the object at t  1.00s .
Solution： (a) By comparing tthe general equation for simple harmonic motion—
x  A cos(t   ) —we see that A  4.00m ,    rad / s . Therefore,
f   2  0.500 Hz , and T  1 f  2.00s .                     (Answer)

(b) Calculate the velocity and acceleration of the object at any time t.
 d                                   
v  dx    (4.00m)sin( t  ) ( t )  (4.00 m / s)sin( t  )
dt                    4 dt                                   4
 d                                      
a  dv  (4.00 m / s) cos( t  ) ( t )  (4.00 2 m / s 2 ) cos( t  )
dt                          4 dt                                     4
(c) Noting that the angles in the trigonometric functions are in radians, we obtain,
at t  1.00s ,
               5
x  (4.00m) cos(  )  (4.00m) cos( )  (4.00m)(0.707)  2.83m
4               4
5
v  (4.00 m / s)sin( )  (4.00 m / s)(0.707)  8.89m / s
4
5
a  (4.00 2 m / s 2 ) cos( )  (4.00 2 m / s 2 )(0.707)  27.9m / s 2
4

Example 4 The thin circle with radius of R, resting on a knife edge O, is in its own
plane for the small swing. Calculate (a) the period of the oscillation, (b) the length of
the pendulum with the same period of oscillation, and (c) cutting off 2 3 of the circle
and putting the center of the remaining arc on the knife. Find the ratio of periods
between the remaining arc and the whole circle.
Solution ： (a) From the period of the physical
o
pendulum, we know
o
I
T  2                                 (1)
Fig.8-2 Example 4.
mgh
where m is the mass of the thin circle.According to the Parallel axis theorem, we
can get


I 0  I 0  moo2  mR 2  mR 2  2mR 2

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8.    Oscillations

Substituting the I 0 into the Eq.(1), yields

2mR 2      2R
T  2        2                                     (2) (Answer)
mgR         g
(b) From the period of the simple pendulum, we get

T0  2                                                (3)
g
Because the period of the simple pendulum is the same as that of the circle, so T  T0 .
Combining the Eq.(2)-(3), we can obtain the length of the simple pendulum
(c) The remain arc whose center is on the knife edge is still the physical pendulum, so
its period is

I
T   2                                                (4)
mgh
From the symmetry property, we know the center of the mass is on the oo . Assuming
m is the mass of the remaining arc. And h  oc is the distance between the center of
the remaining arc and its supporting point o, so the rotational inertia of the remain arc
can be written as

     
I 0  I C  mh2  mR 2  m( R  h) 2  mh2  2mRh          (5)

Substituting the Eq.(5) into the Eq.(4), the period of the remain arc is

2mRh      2R
T   2                2
mgh       g
T
T
No matter how much the circle removed, its period of the oscillation doesn’t change as
long as the knife edge higher than the center of the remaining arc.
Example 5 A ball of mass m1  0.9kg is suspended on from the ceiling with a light
cord of length 0.9m . Another ball of mass m1  0.1kg with velocity v  1.0m / s ,
moving horizontally strikes the ball in completely inelastic collision. Determine the
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8.      Oscillations
kinematics equation after collision of two balls.
Solution ： Considering the balls m1 ,m 2 as a system, the linear momentum is
conserved along the horizontal direction in the completely inelastic collision.
m 2 v2x  (m1  m 2 )vx
m 2v2x
vx               0.1(m / s)
m1  m 2
After collision the m1 and the light line looked as the simple pendulum with the simple
harmonic motion. Assuming its equation is

  A cos(0 t   )                                     (1)

At the initial time     0 , and then we can get
d  x 0.1
0  0,                 angular velocity
dt              0.9
g            9.8
0                     3.3                               (2)
0.9

0 2 0
A  02  (         )      0.0337                         (3)
0     0
From the initial condition, we also know cos   0 , sin    0 A0  1 .
Solving for  , yields

                                                         (4)
2
Combining the Eq.(1)-(4), the the kinematics equation can be written as

  0.0337 cos(3.3t  )                                (Answer)
2

Example 6 A 0.500 kg cube connected to a light spring for which the force constant
is 20.0N / m oscillates on a horizontal, frictionlesstrack. (a) Calculate the total energy
of the system and the maximum speed of the cube if the amplitude of the motion is
3.00cm . (b) What is the velocity of the cube when the displacement is 2.00cm ? (c)
Compute the kinetic and potential energies of the system when the displacement is
2.00cm .
Solution：(a) Using Equation E  K  U , we obtain
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8.   Oscillations
1
E  K  U  (20.0 N / m)(3.00 102 )2  9.00 103 J
2
1 2
When the cube is at x  0 we know that U  0 , and E  mvmax , therefore
2
1 2                             18.0 103
mvmax  9.00 103 J  vmax              0.190m / s                      (Answer)
2                                0.500kg
(b) We can calculate directly:

k 2               20.0
v       ( A  x2 )         [(0.0030) 2  0.0020) 2 ]  0.141m / s (Answer)
m                0.500
The positive and negative signs indicate that the cube could be moving to either the
right or the left at this instant.
(c) Using the result of (b), we find that
1 2 1
K   mv  (0.500kg )(0.141m / s) 2  5.00 103 J
2     2
1 2 1
U  kx  (20.0 N / m)(0.020m) 2  4.00 103 J                  (Answer)
2     2

Problem Solving

1 (2) Fig3 shows block 1 of mass 0.200kg sliding to the right over a frictionless
elevated surface at a speed of 8.00m / s . The block undergoes an elastic collision with
stationary block 2, which is attached to a spring of spring constant 1208.5N / m .
(Assume that the spring does not affect the collision.) After the collision, block 2
oscillates in SHM with a period of 0.140s , 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  4.90m .
What is the value of d?
Solution：The statement that “the spring does not
affect the collision” justifies the use of elastic
collision formulas. We are told the period of SHM
Fig.8-3 Problem 1.
so that we can find the mass of block 2:

m2        kT 2
T  2     m2        0.600kg
k         4 2
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8.   Oscillations
At this point, the rebound speed of block 1 can be found

0.200  0.600
v1 f                  (8.00m / s )  4.00m / s
0.200  0.600

This becomes the initial speed v 0 of the projectile motion of block 1. A variety of
choices for the positive axis directions are possible, and we choose left as the +x
direction and down as the +y direction, in this instance. With the “launch” angle
being zero, we can get

2h          2(4.90)
d  x  x0  v0t  v0          (4.00)          4.00m            (Answer)
g             9.8

2 (4) Two particles oscillate in simple harmonic motion along a common straight line
segment of length A. Each particle has a period of 1.5s , but they differ in phase
 6rad .(a) How far apart are they (in terms of A) 0.50s after the lagging particle
leaves one end of the path? (b) Are they then moving in the same direction, toward
each other, or away from each other?
Solution：(a) Let
A     2 t
x1      cos
2      T
be the coordinate as a function of time for partical 1 and
A      2 t 
x2       cos(      )
2       T   6
be the coordinate as a function of time for particle 2. Here T is the period. Note that
since the range of the motion is A A, the amplitudes are both A 2 . The arguments of
the cosine functions are in radians. Particle 1 is at one end of its path ( x1  A 2 )
when t  0 . Particle 2 is at A 2 when 2 t T   6  0 or t   T 12 . That is,
particle 1 lags particle 2 by one twelfth a period. We want the coordinates of the
particles 0.50s later; that is, at t  0.50s ,
A     (2  0.50s)
x1      cos               0.25 A
2         1.5s
and
A      2  0.50s 
x2      cos(            )  0.43 A
2         1.5s    6

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8.   Oscillations

Their separation at that time is x1  x2  0.25 A  0.43 A  0.18 A .         (Answer)

(b) The velocities of the particles are given by
dx1  A     2
v1           sin( )
dt   T      T
and
dx2  A     2 
v2            sin(  )                        (Answer)
dt   T      T 6
We evaluate these expressions for t  0.50s and find they are both negative-valued,
indicating that the particles are moving in the same direction.

3 (7) In Fig.3, two blocks ( m  1.0kg and M  10kg ) and a spring ( k  200N / m ) are
arranged on a horizontal, frictionless surface. The coefficient of static friction between
the two blocks is 0.40 . What amplitude of simple harmonic motion of the
springblocks system puts the smaller block on the
verge of splipping over the large block?
Solution： The key idea is that the smaller block
gives a force which acted on the larger block, that is

f   mg  0.4  0.1kg 10m / s 2  0.4 N                  Fig.8-4 Problem 3.

We know the sustem oscillate in SHM, and when the smaller block slipping over the
large block, we have
kA   mg  mg

M       m
Solving for d and substituting known data, we can obtain the amplitude

 (m  M ) g          0.4 11kg 10m / s 2
k                    200 N / m
4 (8) In Fig.5, two spring are joined and connected to a block of mass 0.245kg
that is set oscillating over a frictionless floor. The spring each have spring constant
k  6430N / m . What is the frequency of the
oscillator?
Solution：We wish to find the effective spring
constant for the combination of springs shown in
Fig. 5. We do this by finding the magnitude F of
Fig.8-5 Problem 4.
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8.   Oscillations
the force exerted on the mass when the total elongation of the springs is Δx. Then
keff  F  x . Suppose the left-hand spring is elongated by  xl and the right-hand
spring is elongated by  xr The left-hand spring exerts a force of magnitude
k xl on the right-hand spring and the right-hand spring exerts a force of
magnitude k xr on the left-hand spring. By Newton’s third law these must be equal,
so xl  xr . The two elongations must be the same and the total elongation is twice
the elongation of either spring: x  2xl . The left-hand spring exerts a force on the
block and its magnitude is F  k xl .

Thus keff  k xl 2xr  k 2 . The block behaves as if it were subject to the force

of a single spring, with spring constant k 2 . To find the frequency of its motion

1    keff
replace keff in f                   with k 2 to obtain
2     m

1       k   1         6430
f                                   18.2 Hz              (Answer)
2      2m 2        2  0.245

5(9) In Fig. 6, a block weight 14.0N , which can slide without friction on an incline
at   40.00 , is connected to the top of the incline by a massless spring of unstreched
length 0.450m and spring constant 120N / m . (a) How far from the top of the
incline is the block’s equilibrium point? (b) If the block is pulled slightly down the
incline and released, what is the period of the
resulting oscillations?
(a)
Solution： The key idea is that the block stays in
equilibrium is the block’s equilibrium point, so we
have
Fig.8-6 Problem 5.
N  mg sin 
Substitute the known data then yield:
N  9.002N
We have N  kx, x  0.075m

So the distance from the top of the incline to the block’s equilibrium point is

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8.   Oscillations

x  x  L  0.075m  0.45m  0.525m

(b) If the block is pulled slightly down the incline and released, because the oscillation

is SHM, so we have T  2 m k

substituting the known data , yields

14 N
T  2 m k  2 G kg  2  3.14                                        0.686 s (Answer)
120 N / m  9.8m / s 2

6 (10) In Fig.7, block 2 of mass 2.0kg oscillates on the end of a spring in SHM with
a period of 20ms . The position of the block is given by x  (1.0cm)cos(t   2).
Block 1 of mass 4.0kg slides toward block 2
with a velocity of magnitude 6.0m / s , directed
along the spring’s length. The two blocks udergo a
completely inelastic collision at time t  5.0ms .
(The duration of the collision is much less than the          Fig.8-7 Problem 6.
period of motion.) What is the amplitude of the SHM after the collision?
Solution：We note that the spring constant is

4 2 m1
k     2
 1.97 105 N / m
T
It is important to determine where in its simple harmonic motion (which “phase” of
its motion) block 2 iswhen the impact occurs. Since   2 T and the given value
of t (when the collision takes place) is one-fourth of T, then t  2 and the
location then of block 2 is x  (1.0cm)cos(t   2). where    2 which gives
x  1.0m . This means block 2 is at a turning point in its motion (and thus has zero
speed right before the impact occurs).
Since the two blocks stick together in the process, we use momentum conservation and
obtain

m1v1  (m1  m2 )v

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8.   Oscillations

(4.0kg )(6.0m / s)
Solving for v , yields v                         4.0m / s
6.0kg
And the system has kinetic energy and potential energy, respectively
1               1
K  (m1  m2 )v 2  (6.0kg )(4.0m / s) 2  48 J
2               2
1 2 1
U  kx  (1.97 105 N / m)(0.010m) 2  10 J
2      2
meaning the total mechanical energy in the system at this stage is approximately
58J . When the system reaches its new turning point (at the new amplitude X ) then
this amount must equal its (maximum) potential energy there:
1
(1.97 105 ) X 2
2
Therefore, we find

X  2(58) /(1.97 105 )  0.024m                (Answer)

7 (11) A block of mass M  5.4kg , at rest on a horizontal frictionless table, is
attached to a rigid support by a spring of constant k=6000N/m. A bullet of mass
m  9.5 g and velocity v of magnitude 630m / s strikes and is embedded in the
block (Fig 8). Assuming the compression of the spring is negnigible untile the bullet is
embedded, determine (a) the speed of the block immediately after the collision and (b)
the amplitude of the resulting simple harmonic motion.
Solution：(a) When the bullet strikes the block which is on the horizontal frictionless
table, the momentum of the system1 composed
with the bullet and the block is conservative. So
we can obtain the speed of the block immediately
after the collision

mv  (m  M )v1                              Fig.8-8 Problem 7.

So the speed of the block immediately after the collision is

mv         9.5
v1                        1.11m / s                  (Answer)
(m  M ) 9.5  5400
(b) Then we can get the kinetic energy of the system1

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8.    Oscillations

1            1             ( mv) 2     m2v2
EK  (m  M )v1  ( m  M )
2

2            2          ( m  M ) 2 2( m  M )
Next, we consider the bullet、the block and the spring as the system2, the mechanic
energy of the system2 is conserved. When the kinetic energy is zero, the spring
potential energy is max and we can get the amplitude of the resulting SHM

1 2             m2v 2
UK      kxm  EK 
2            2( m  M )
Then we can get amplitude of the resulting simple harmonic motion

1                            1
xm  mv                0.0095  630                     3.32 102 m  3.32cm
(m  M )k                (0.0095  5.4)6000

8 (16) In Fig.9, a stick of length L =1.85m oscillates as a physical pendulum. (a)What
value of distance x between the stick’s center of mass and its pivot point O gives the
least period? (b)What is that least period?
Solution ：        The period of the physical pendulum

I
is T  2       , so the key idea here is what is I. And the
mgx
1
rotational inertia of the pendulum is I       mL2  mx 2
12
1
mL2  mx 2
L2    x
So T  2 12             2      
mgx            12 gx g                                Fig.8-9 Problem 8.

L2    x
When           , we can get the minimum value of the time
12 gx g
2                 2
So                       x L             1.85           0.534m            (Answer)
12                12
(b) From (a) we know

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8.   Oscillations

3L          3 1.85
T  2          2                 2.07 s             (Answer)
3g       3  9.8m / s 2

9 (18) The angle of the pendulum of Fig. 10 is given by   m cos[(4.44rad / s )t   ].
If at t  0 ,   0.04rad and d dt  0.200 rad / s , what are (a) the phase
constant  and (b) the maximum angle  m ? (Hint: Don't
confuse the rate d dt at which  changes with the w of
the SHM.)
Solution： As the angle of the pendulum is given by

   m cos[(4.44rad / s)t   ] , we can differentiate
 with respect to t then we can get
d
 4.44 m sin[(4.44rad / s)t   ]
dt
Substituting the data to the above equation, we can get             Fig.8-10 Problem 9.

 m cos   0.04,  4.44 m sin   0.2

(a) tan   1.126,   49.24
0

0.04
cos 

10 (19) In Fig.11a, a metal plate is mounted on an axle through its center of mass. A
spring with k  200N / m connects a wall with a point on the rim a
distance r  2.5cm from the center of mass.
Initially the spring is at its rest length. If the
plate is rotated by 7 0 and released, it rotates
about the axle in SHM, with its angular position
given by Fig.11b.What is the rotational inertia of
the plate about its center of mass?
Fig.8-11 Problem 10.
Solution：We note that the initial angle is  0  7  0.122rad (though it turns out
0

this value will cancel in later calculations). If we approximate the initial stretch of the
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8.   Oscillations
spring as the arclength that the corresponding point on the plate has moved through

( x  r 0 , where r  0.025m ) then the initial potential energy is approximately,

1 2
kx  2000  (0.025  0.122)2  0.0093J
2
1
This should equal to the kinetic energy of the plate (    I  2 where this m is the
12
maximum angular speed of the plate, not the angular frequency  ). Noting that the

maximum angular speed of the plate is m   0 , where   2 T                      with

T  20ms  0.02m . as determined from the graph, then we can find the rotational
1
inertial from    I  2  0.0093J , thus
12
I  1.3 105 kg  m 2 .                      (Answer)

11 (20) A pendulum is formed by pivoting a long thin rod about a point on the rod.
In a series of experiments, the period is measured as a function of the distance d
between the pivot point and the rod’s center. (a) If the rod’s length is L  2.20m and
its mass is m  22.1g , what is the minimum period? (b) If d is chosen to minimize
the period and then L is increased, does the period increase, decrease, or remain the
same? (c) If, instead, m is increased without L increasing, does the period increase,
decrease, or remain the same?
Solution：A key idea here is that the system is a physical pendulum, the equation of
period for physical pendulum can be used.
(a) As we know the thin rod about axis through center perpendicular to the length is
1 12 ML2 . And according to the parallel-axis theroem, the rotational inetia of the
thin rod about pivot point is
1 2
I  I com  md 2         ml  md 2
12
And the period of the physical pendulum is

I       I com  md 2       l2    d
T  2      2               2      
mgd          mgd           12 gd g

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8.   Oscillations

l2   d               l
When       , that is d      , the period has its minimum value
12 gd g              2 3

3l                   3  2.20
Tmin  2          2  3.14                  2.26s (1)        (Answer)
3g                   3  9.80
(b) If L is increased, we can esily draw a conclusion based on Eq.(1) that the period
(c) Because the expression of Tmin is absent from the mass of the thin rod, the period
would remain the same value when m is increased.                            (Answer)

12 (21) In Fig. 12, a 2.50kg disk of diameter D  42.0cm is supported by a rod of
length L  76.0cm and negligible mass that is pivoted at
its end. (a)With the massless torsion spring unconnected,
what is the period of oscillation? (b)With the torsion spring
connected, the rod is vertical at equilibrium. What is the
torsion constant of the spring if the period of oscillation has
been decreased by 0.500s ?
Solution： The key idea is that the system can be seen as a
simple pendulum and the torsion spring would decrease the
period when connected.                                         Fig.8-12 Problem 12.
(a) With the massless torsion spring unconnected, the torque arm of the pendulum is
D            42.0
L0  L        76.0cm       cm  97.0cm
2             2
So the oscillation’s period is

L0         0.97
T  2         2              2.00s               (Answer)
g       9.80m / s 2
(b) Now we have two “restoring torques” acting in tandem to pull the pendulum back
to the vertical position when it is displaced. The magnitude of the net torque is
( Mgh   ) , where the small angle approximation ( sin   in radians). From the
equation of the period of torsion pendulum, making the appropriate adjustment to the
period formula, we have

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8.   Oscillations

IP
T   2
Mgh  

where Mgh   is the composite torsion constant.

The problem statement requires T  T   0.05s . Thus T   (2.00  0.50) s  1.50s .

Solving for  , yields

4 2
        I P  Mgh  18.5 N .m / rad        (Answer)
T 2

“
13 (23) A 1000kg car carrying four 82kg people travels over a washboard” dirt
road with corrugations 4.0m apart. The car bounces with maximum amplitude when its
speed is 16km / h . When the car stops, and the people get out, by how much does the
car body rise on its suspension?
Solution： With M  1000kg and m  82kg , we can write the period of the system

M  4m
T  2
k

If d  4.0m is the distance traveled (at constant car speed v ) between impulses,
then we may write T  d v , in which case the above equation may be solved for the
spring constant

d      M  4m                 2
 2         k  ( M  4m)( ) 2
v        k                     d

Before the people got out, the equilibrium compression is xi 
( M  4 m) g
k , and

afterward it is x f 
Mg                       v  16000       4.44m / s , we find
k . Therefore, with           3600
the rise of the car body on its suspension is
4mg    4mg    d
xi  x f               ( )2  0.050m                    (Answer)
k    M  4m 2

14 (24)     Figure13 five the position of
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8.   Oscillations
a 20g block oscillating in SHM on the end of a spring. What are (a) the maximum
kinetic energy of the block and (b) the number of times per second that maximum is
reached? (Hint: Measuring a slope will probably not be very accurate. Can you think
of another approach?)
Solution： The key idea is that the motion of the block is SHM.
(a) From the figure, we can get the period T
and amplitude of the system, they are
Fig.8-13 Problem 14.
T  40ms, xm  7cm

so the angular frequency is
2
       50 rad / s
T
The total mechanical energy of the SHM is
1         1
E  m 2 xm   0.02kg  (50 rad / s) 2  (0.07m) 2  1.21J
2

2         2
When the block is at its equilibrium point, it has a maximum kinetic energy, and it
equals the total mechanical energy, that is

Em  E  1.21J                                (Answer)


(b) Because the frquency is f           25Hz , in one period, the block passes
2
through the equilibrium point twice. So the number of times per second that maximum

15 (29) In Fig. 14, three 10000kg ore cars are held at rest on a mine railway using
a cable that is parallel to the rails, which are inclined at angle   300 . The cable
stretches 15cm just before the coupling between the two lower cars breaks,
detaching the lowest car. Assuming that the cable
obeys Hooker’s law, find the (a) frequency and (b)
amplitude of the resulting oscillations of the
remaining two cars.
Solution： The magnitude of the downhill
component of the gravitational force acting on
Fig.14 Problem 15.
each ore car is

Fx  mg sin   (10000kg )(9.8m / s 2 ) sin 
- 208 -
8.   Oscillations
where   30
0
(and it is important to have the calculator in degrees mode during this
problem). We are told that a downhill pull of 3 x causes the cable to stretch
x  0.15m . Since the cable is expected to obey Hook’s law, its spring constant is
3F
k  x  9.8 105 N / m .
x
(a) Noting that the oscillating mass is that of two of the cars, we can get the frequency

1    9.8 105
2    20000kg
(b) The difference between the equilibrium positions of the end of the cable when
supporting two as opposed to three cars is
3Fx  2 Fx
k

16 (32)    A 0.10 kg block oscillates back and forth along a straight line on a
frictionless horizontal surface, Its displacement from the origin is given by

(a)What is the oscillation frequency? (b) What is the maximum speed acquired by the
block (c) At what value of x does this occur? (d) What is the magnitude of the
maximum acceleration of the block? (e)At what value of x does this occur? (f)What
force, applied to the block by the spring, results in the given oscillation?
Solution： (a) A key idea here is that the motion of the block is SHM. From equation
given by the title, we know that
 10
f          1.6 Hz
2 2
dx
dt                                                                 (1)
 (100cm / s ) sin[(10rad / s )t   2rad ]

dv
a      (100cm / s ) cos[(10rad / s )t   2rad ](10rad / s )
 (1000cm / s ) cos[(10rad / s )t   2rad ]
2

- 209 -
8.   Oscillations

(b) From the Eq.(1), the maximum speed is vmax  100cm / s                     (Answer)

(c) When the speed is max, the value of x is zero.                             (Answer)
2
(d) The maximum acceleration of the block is 1000cm / s , given by the Eq.(2).

(e) When the acceleration is max, the value of x is 10cm .                  (Answer)
(f) According to the Newton’s second law, we can get the force applied to the block

F  ma  0.1kg (10m / s 2 ) cos[(10rad / s)t   2rad ]

17 (33) In Fig.15, a solid cylinder attached to a horizontal spring ( k  3.00N / m )
rolls without slipping along a horizontal surface. If the system is released from rest
when the spring is stretched by 0.250m , find (a) the translational kinetic energy and (b)
the rotational kinetic energy of the cylinder as it passes through the equilibrium
position. (c) Show that under these conditions the cylinder’s center lf mass executes
simple harmonic motion with period

3M
T  2
2k
Where M is the cylinder mass. (Hint: Find the time derivative of the total mechanical
energy.)
Solution：(a) The potential energy at the turning
point is equal (in the absence of friction) to the
total kinetic energy (translational plus rotational)
as it passes through the equilibrium position:
Fig.15 Problem 17.

1 2 1          1 2         1      1 1       v
kxm  Mvcom  I com 2  Mvcom  ( MR 2 )( com ) 2
2                   2

2       2       2          2      2 2         R
1       1          3
 Mvcom  Mvcom  Mvcm
2         2        2

2       4          4

which leads to Mvcom  2kxm 3  0.125 J . The translational kinetic energy is
2        2

- 210 -
8.    Oscillations
1
Mvcom  kxm 3  0.0625 J
2       2
2
(b) And the rotational kinetic energy is
1
Mvcom  kxm 6  3.13 102 J
2       2
4
(c) In this part, we use vcom to denote the speed at any instant (and not just the

maximum speed as we had done in the previous parts). Since the energy is constant,
then
dE d 3         d 1        3
 ( Mvcom )  ( kx 2 )  Mvcom acom  kxvcom  0
2

dt dt 4        dt 2       2
2k
acom  (      )x
3M
We see that     2k 3M for this system. Since   2 T , we obtain the desired
result:

3M
2k

18 (34) An engineer has an odd-shaped 10kg object and needs to find its rotational
inertia about an axis through its center of mass. The object is supported on a wire
stretched along the desired axis. The wire has a torsion constant   0.50N m If
this torsion pendulum oscillates through 20 cycles in 50s , what is the rotational
inertia of the object?
Solution： The key idea is that T  2 I  . So we substitute the data into the
equation and then get the rotational inertia of the object

50 s
T  2 I  
20
2.5
I  ( ) 2  0.5  0.079kg  m 2
2

19 (35) A block weighing 10.0N is attached to the lower end of a vertical spring

- 211 -
8.   Oscillations
( K  200.0N / m ), the other end of which is attached to a ceiling. The block oscillates
vertically and has a kinetic energy of 2.00J as it passes through the point at which
the spring is unstretched. (a)What is the period of the oscillation? (b) Use the law of
conservation of energy to determine the maximum distance the block moves both
above and below the point at which the spring is unstretched. (These are not
necessarily the same.) (c) What is the amplitude of the oscillation? (d) What is the
maximum kinetic energy of the block as it oscillates?
Solution： The key idea is that the motion of the block is SHM, so we can get

(a) T  2 m k  2 10 9.8  200.0  0.45s                                    (Answer)

(b) For a vertical spring, the distance between the unstretched length and the
equilibrium length (with a mass m attached) is
mg     10 N
x                     0.05m
k   200.0 N / m
During simple harmonic motion, the convention is to establish x  0 at the
equilibrium position (the middle level for the oscillation) and to write the total energy
1
E  K  U  K  kx 2
2
Thus, as the block passes through the unstretched position, the energy is
1              1
E  K  kx 2  2.0 J  (200 N / m) 2 (0.05m) 2  2.25 J
2              2
At its topmost and bottommost points of oscillation, the energy (using this convention)
1
kxm 2 .
is all elastic potential
2
1
kxm 2  2.25 J  xm  0.15m            (Answer)
2
This gives the amplitude of oscillation as 0.15m , We add (or subtract) the
0.05m value found above and obtain 0.10m for the top-most position and 0.20m
for the bottom-most position.

(c) As noted in part (b), xm  0.15m                                          (Answer)

(d) The maximum kinetic energy equals the maximum potential energy (found in part
(b))and is equal to 2.25J .

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8.   Oscillations
20 (38) In Fig. 16, a 2500kg demolition ball swings from the end of a crane. The
length of the swinging segment of cable is 17m . (a)
Find the period of the swinginng, assuming that the
system can be treated as a simple pendulum. (b) Does
the period depend on the ball’s mass?
Solution：(a) Because the system can be treated as a
simple pendulum, the period is
Fig.16 Problem 20.
L         17m
T  2       2              8.3s                        (Answer)
g      9.80m / s 2

I
(b) Plugging I  mL into T  2
2
, we see that the mass m cancels out.
mgL
Thus, the characteristics (such as the period) of the periodic motion do not depend on

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