Download Chapter 15: Oscillations 15-23 THINK The maximum force that can

Chapter 15: Oscillations
15-23
THINK The maximum force that can be exerted by the surface must be less than the
static frictional force or else the block will not follow the surface in its motion.
EXPRESS The static frictional force is given by f s  s FN , where µs is the coefficient
of static friction and FN is the normal force exerted by the surface on the block. Since the
block does not accelerate vertically, we know that FN = mg, where m is the mass of the
block. If the block follows the table and moves in simple harmonic motion, the
magnitude of the maximum force exerted on it is given by
F = mam = m2xm = m(2f)2xm,
where am is the magnitude of the maximum acceleration,  is the angular frequency, and
f is the frequency. The relationship  = 2f was used to obtain the last form.
ANALYZE We substitute F = m(2f)2xm and FN = mg into F  µsFN to obtain m(2f)2xm
 µsmg. The largest amplitude for which the block does not slip is
xm
b0.50gc9.8 m / s h 0.031 m.
=
=
b2f g b2  2.0 Hzg
sg
2
2
2
LEARN A larger amplitude would require a larger force at the end points of the motion.
The block slips if the surface cannot supply a larger force.
15-39
THINK The balance wheel in the watch undergoes angular simple harmonic oscillation.
From the amplitude and period, we can calculate the corresponding angular velocity and
angular acceleration.
EXPRESS We take the angular displacement of the wheel to be t = m cos(2t/T),
where m is the amplitude and T is the period. We differentiate with respect to time to
find the angular velocity:
 = d/dt = –(2/T)msin(2t/T).
The symbol  is used for the angular velocity of the wheel so it is not confused with the
angular frequency.
ANALYZE (a) The maximum angular velocity is
m 
2 m  2  rad 

 39.5 rad/s.
T
0.500 s
(b) When  = /2, then /m = 1/2, cos(2t/T) = 1/2, and
sin  2 t T   1  cos2  2 t T   1  1 2   3 2
2
where the trigonometric identity cos2+sin2= 1 is used. Thus,
=
F I F
HKH
Ia
K
2
2 t
2
 msin
=
 rad
T
T
0.500 s
3I
fF
G
H2 J
K= 34.2 rad / s.
During another portion of the cycle its angular speed is +34.2 rad/s when its angular
displacement is /2 rad.
(c) The angular acceleration is
d 2
 2 
 2 
 
  m cos  2 t / T    
 .
2
dt
 T 
 T 
2

When  = /4,
2
 2    
2
  
   = 124 rad/s ,
 0.500 s   4 
2
or |  |  124 rad/s 2 .
LEARN The angular displacement, angular velocity and angular acceleration as a
function of time are plotted below.
15-59
THINK In the presence of a damping force, the amplitude of oscillation of the massspring system decreases with time.
EXPRESS As discussed in 15-8, when a damping force is present, we have
x(t )  xm e bt / 2 m cos( t   )
where b is the damping constant and the angular frequency is given by   
k
b2
.

m 4m 2
ANALYZE (a) We want to solve e–bt/2m = 1/3 for t. We take the natural logarithm of both
sides to obtain –bt/2m = ln(1/3). Therefore,
t = –(2m/b) ln(1/3) = (2m/b) ln 3.
Thus,
t
2 1.50 kg 
0.230 kg/s
ln 3  14.3 s.
(b) The angular frequency is
 
a
f
k
b2
8.00 N / m 0.230 kg / s



2
2
m 4m
1.50 kg
4 1.50 kg
a f
2
 2.31 rad / s.
The period is T = 2/´ = (2)/(2.31 rad/s) = 2.72 s and the number of oscillations is
t/T = (14.3 s)/(2.72 s) = 5.27.
LEARN The displacement x(t) as a function of time is shown below. The amplitude,
xm e  bt / 2 m , decreases exponentially with time.