Download Ch 14 HW Day 2 p 455 – 464

Ch 14 HW Day 2 p 455 – 464, #’s 33-39 (ODDS), 40*, 41-59* (ODDS, omit #53)
Simple Harmonic Motion and Circular Motion
33 •
Picture the Problem We can find the period of the motion from
the time required for the particle to travel completely around the
circle. The frequency of the motion is the reciprocal of its period
and the x-component of the particle’s position is given
by x
 A cost    .
(b) Use the definition of
speed to find the period
of the motion:
(a) Because the
frequency and the
period are reciprocals
of each other:
(c) Express the x
component of the
position of the particle:
Assuming that the particle
is on the positive x axis at
time t = 0:
Substitute for A, , and  to
obtain:
T = 2/;
Or, use 2r/v = T, subst, 
2(0.4m)/.8m/s  s
Subst will give you 2/s
2s-1 (use for )
T
2r 2 0.4 m 

 3.14 s
v
0.8 m/s
f 
1
1

 0.318 Hz
T 3.14 s
x  A cost   
A  Acos  
  cos 1 1  0
x  A cos2ft 

40 cm cos2 s1 t 
Energy in Simple Harmonic Motion
35 •
Picture the Problem The total energy of the object is given by
Etot  12 kA2 , where A is the amplitude of the object’s motion.
Express the total
energy of the system:
Etot  12 kA2
Substitute numerical
values and evaluate
Etot:
Etot 
1
2
4.5 kN/m0.1m2 
22.5 J
37 •
Picture the Problem The total mechanical energy of the
oscillating object can be expressed in terms of its kinetic energy
2
1
as it passes through its equilibrium position: Etot  2 mvmax . Its
total energy is also given by Etot  2 kA . (E is conserved, so
when v is a max, K is a max and U is a min, or 0.) We can
equate these expressions to obtain an expression for A.
1
2
(a) Express the total
mechanical energy of
the object in terms of
its maximum kinetic
energy:
2
E  12 mvmax
Substitute numerical
values and evaluate E:
E
(b) Express the total
energy of the object in
terms of the amplitude
of its motion:
Etot  12 kA2
1
2
1.5 kg0.7 m/s 2 
0.368 J
Solve for A:
Substitute numerical
values and evaluate A:
A
2 Etot
k
A
20.368 J 
 3.84 cm
500 N/m
39 •
Picture the Problem The total energy of the object is given by
Etot  12 kA2 . We can solve this equation for the force constant k
and substitute the numerical data to determine its value.
Express the total
energy of the oscillator
as a function of the
amplitude of its
motion:
Etot  12 kA2
Solve for k:
k
2 E tot
A2
Substitute numerical
values and evaluate k:
k
21.4 J 
 1.38 kN/m
2
0.045 m 
*40 ••
Picture the Problem The total energy of the object is given, in
2
. We can
terms of its maximum kinetic energy by Etot  12 mvmax
express vmax in terms of A and  and, in turn, express  in
terms of amax to obtain an expression for Etot in terms of amax.
Express the total
energy of the object in
terms of its maximum
kinetic energy:
2
Etot  12 mvmax
Relate the maximum
speed of the object to
its angular frequency:
Substitute to obtain:
Relate the maximum
acceleration of the
object to its angular
frequency:
Substitute and simplify to
obtain:
Substitute numerical
values and evaluate
Etot:
vmax  A
(v is the first derivative of x; sin and cos
have max magnitudes at 1, so those
portions of the respective eqs  1, 
vmax = A∙1, or just A. This
generalization is true, respectively, for
amax, as well, so amax, which = d2x/dt2, =
A2.)
Etot  12 m A   12 mA2 2
2
amax  A 2
or
2 
amax
A
Etot  12 mA2
amax 1
 2 mAamax
A

Etot  12 3 kg 0.08 m  3.50 m/s 2

 0.420 J
Springs
41 •
Picture the Problem The frequency of the object’s motion is given by
f 
1
2
k m.
(f = 1/2  ω and  =
k
m
). Its period is the reciprocal of its
frequency. The maximum velocity and acceleration of an object
executing simple harmonic motion are vmax  A and amax  A 2 ,
respectively.
(a) The frequency of
the motion is given by:
f 
1
2
k
m
Substitute numerical
values and evaluate f:
f 
1
2
4.5 kN/m
 6.89 Hz
2.4 kg
Remember to put 2 in parentheses, or
you’ll get 68.0 as your answer!
1
1

 0.145
f 6.89 s 1
(b) The period of the
motion to is the
reciprocal of its
frequency:
(c) Because the object
is released from rest
after the spring to
which it is attached is
stretched 10 cm:
T
(d) Express the object’s
maximum speed:
vmax  A  2fA
Substitute numerical
values and evaluate
vmax:
vmax  2 6.89 s 1 0.1m  4.33 m/s
(e) Express the object’s
maximum acceleration:
amax  A 2  vmax  2fvmax
Substitute numerical
values and evaluate
amax:
amax  2 6.89 s 1 4.33 m/s 
(f) The object first
reaches its equilibrium
when:
t  14 T  14 0.145 s  0.0363 s
A  0.100 m




 187 m/s 2
See #41 scanned image
Because the resultant
force acting on the
object as it passes
through its equilibrium
point is zero, the
acceleration of the
a 0
object is:
43 •
Picture the Problem The angular frequency, in terms of the
force constant of the spring and the mass of the oscillating
object, is given by   k m . The period of the motion is the
reciprocal of its frequency. The maximum velocity and
acceleration of an object executing simple harmonic motion are
2
2
vmax  A and amax  A , respectively.
(a) Relate the angular
frequency of the motion
to the force constant of
the spring:
2 
k
m
or
k  m 2  4 2 f 2 m

 3 kg 
Substitute numerical
values to obtain:
k  4 2 2.4 s 1
(b) Relate the period of
the motion to its
frequency:
(c) Express the
maximum speed of the
object:
T
Substitute numerical
values and evaluate
vmax:
vmax  2 2.4 s1 0.1m  1.51m/s
(d) Express the
maximum acceleration
of the object:
amax  A 2  4 2 f 2 A
Substitute numerical
values and evaluate
amax:
amax  4 2 2.4 s 1
2
682 N/m
1
1

 0.417 s
f 2.4 s 1
vmax  A  2fA



 0.1m 
2
22.7 m/s 2
45 •
Picture the Problem We can relate the force constant k to the
maximum acceleration by eliminating 2 between
 2  k m and amax  A 2 . We can also express the frequency f of the
motion by substituting mamax/A for k in
k
m
(a) Relate the angular
frequency of the motion
to the force constant and
the mass of the
oscillator:
2 
Relate the object’s
maximum acceleration
to its angular frequency
and amplitude and solve
for the square of the
angular frequency:
Substitute to obtain:
amax  A 2
f 
1
2
k
.
m
or k   2m
or
2 
amax
A
(1)
k
Substitute numerical
values and evaluate k:
(b) Replace  in
equation (1) by 2f and
solve for f to obtain:
Substitute numerical
values and evaluate f:
k
(c) The period of the motion
is the reciprocal of its
frequency:
mamax
A
4.5 kg 26 m/s 2  
3.8 102 m
3.08 kN/m
f 
1
2
amax
A
f 
1
2
26 m/s 2
 4.16 Hz
3.8 102 m
T
1
1

 0.240 s
f 4.16 s 1
47 ••
Picture the Problem The maximum speed of the block is given
by vmax  A and the angular frequency of the motion
is   k m  5.48 rad/s . We’ll assume that the position of the
block is given by x  A cos t and solve for t for x = 4 cm
and x = 0. We can use these values for t to find the time for
the block to travel from x = 4 cm to its equilibrium position.
(a) Express the
maximum speed of the
block as a function of
the system’s angular
frequency:
Substitute numerical
values and evaluate
vmax:
(b) Assuming
that x  A cos  t, evaluate
t for x = 4 cm = A/2:
Evaluate v for t  
3:
vmax  A
vmax  0.08 m 5.48 rad/s 
 0.438 m/s
A
1 
 A cos t  t  cos 1 
2
2 3
I would write this as:
4 = 8 cos  t. So,  t = cos -1 (1/2).
Then, they subst that into the equation
for v.
v  vmax sin t  0.438 m/s sin
 0.438 m/s 

3
3
 0.379 m/s
2
Express a as a function
of vmax and :
a  A 2 cos t  vmax  cos t
Substitute numerical
values and evaluate a:
a  0.438 m/s 5.48 rad/s cos

3
 1.20 m/s 2
(c) Evaluate t for x = 0:
0  A cos t  t  cos 1 0 

2
Let t = time to go from
t   3 to
t   2 . Then:
Solve for and evaluate t:
t 
t 

2


3


6



 95.5 ms
6 65.48 rad/s 
Or, you could find the entire period,
and divide it by 4!
49 ••
Picture the Problem Let the system include the object and the
spring. Then, the net external force acting on the system is
zero. Choose Ei = 0 and apply the conservation of mechanical
energy to the system.
2
Express the period of
the motion in terms of
its angular frequency:
T
Apply conservation of
energy to the system:
Ei  Ef or 0  U g  U spring
Substitute for Ug and
Uspring:
Solve for 2 = k/m:
0  mgx  12 k x 
Substitute numerical
values and evaluate 2:
Substitute in equation (1) to
obtain:
(1)

2
2 
k 2g

m x
2 
29.81m/s 2 
 574 rad/s 2
2
3.42 10 m
T
2π
574 rad/s 2
 0.262 s
51 ••
Picture the Problem The stunt woman’s kinetic energy, after 2
s of flight, is K2 s  12 mv22s . We can evaluate this quantity as soon as
we know how fast she is moving after two seconds. Because her
motion is oscillatory, her velocity as a function of time is
vt    A sin t   . We can find the amplitude of her motion from
her distance of fall and the angular frequency of her motion by
applying conservation of energy to her fall to the ground.
Express the kinetic
energy of the stunt
woman when she has
fallen for 2 s:
Express her velocity as
a function of time:
Letting Ei = 0, use
conservation of energy
to find the force
constant of the elastic
band:
K 2s  12 mv22s
(1)
vt    A sin t   
where  = 0 (she starts from
rest with positive
displacement) and
A  12 192 m  96 m
(2)
vt   96 m sin t 
0  U g  U elastic
or
0  mgh  12 kh2  0
Solve for k:
k
2mg
h
Substitute numerical
values and evaluate k:
k
260 kg 9.81m/s 2 
 6.13 N/m
192 m
Express the angular
frequency of her
motion:
Substitute numerical
values and evaluate :

k
m

6.13 N/m
 0.320 rad/s
60 kg
Substitute in equation
(2) to obtain:
Evaluate v(2 s):
vt   96 m0.320 rad/s 
 sin 0.320 rad/s t 
 30.7 m/s sin 0.320 rad/s t 
v2 s   30.7 m/s sin 0.320 rad/s 2 s 
 18.3 m/s
Substitute in equation
(1) and evaluate K(2 s):
K 2 s 
1
2
60 kg18.3 m/s 2 
10.1kJ
55 ••
Picture the Problem The maximum height above the floor to
which the object rises is the sum of its initial distance from the
floor and the amplitude of its motion. We can find the
amplitude of its motion by relating it to the object’s maximum
speed. Because the object initially travels downward, it will be
three-fourths of the way through its cycle when it first reaches
its maximum height. We can find the minimum initial speed
the object would need to be given in order for the spring to
become uncompressed by applying conservation of energy.
(a) Relate h, the
maximum height above
the floor to which the
object rises, to the
amplitude of its
motion:
h = A + 5.0 cm
Relate the maximum
speed of the object to
the angular frequency
and amplitude of its
motion and solve for
the amplitude:
vmax  A
Using its definition,
express and evaluate
the force constant of
the spring:
Substitute numerical
values in equation (2)
and evaluate A:
Substitute in equation (1) to
obtain:
(1)
or
A  vmax
k
m
k
(2)
mg 2 kg 9.81 m/s 2 

 654 N/m
y
0.03 m
A  0.3 m/s
2 kg
 1.66 cm
654 N/m
h  1.66 cm  5.00 cm  6.66 cm
(b) Express the time
required for the object
to reach its maximum
height the first time:
t  34 T
Express the period of
the motion:
T  2
m
k
Substitute numerical
values and evaluate T:
T  2
2 kg
 0.347 s
654 N/m
Substitute to obtain:
t
(c) Because h < 8.0 cm:
Using conservation of
energy and letting Ug
be zero 5 cm above the
floor, relate the height
to which the object
rises, y, to its initial
kinetic energy:
Because y  L  yi :
3
4
0.347 s 
0.261s
the spring is never uncompress ed.
K  U g  U s  0
or, because Kf = Ui = 0,
2
2
1
1
2 mvi  mgy  2 k y 
2
 12 k L  yi   0
1
2
mvi2  mgy  12 k y   12 k y   0
2
2
and
1
2
Solve for and evaluate vi
for
y = 3 cm:
mvi2  mgy  0


vi  2 gy  2 9.81 m/s 2 3 cm 
 0.767 m/s
i.e., the minimum initial
velocity that must be given
to the object for the spring
to be uncompressed at some
time is 0.767 m/s
Energy of an Object on a Vertical Spring
57 ••
Picture the Problem Let the origin of our coordinate system be
at y0, where y0 is the equilibrium position of the object and let
Ug = 0 at this location. Because Fnet = 0 at equilibrium, the
extension of the spring is then y0 = mg/k, and the potential
energy stored in the spring is U s  12 ky02 . A further extension of
the spring by an amount y increases Us to
2
2
2
2
2
1
1
1
1
1
2 k  y  y0   2 ky  kyy0  2 ky0  2 ky  mgy  2 ky0 . Consequently, if we set
U = Ug + Us = 0, a further extension of the spring by y increases
Us by ½ky2 + mgy while decreasing Ug by mgy. Therefore, if U =
0 at the equilibrium position, the change in U is given by
2
1
2 k  y ' , where y = y  y0.
(a) Express the total
energy of the system:
E  12 kA2
Substitute numerical
values and evaluate E:
E
(b) Express and
evaluate Ug when the
object is at its
maximum downward
displacement:
(c) When the object is
at its maximum
downward
displacement:
U g  mgA
(d) The object has its
maximum kinetic
energy when it is
passing through its
equilibrium position:
K max  12 kA2 
1
2
600 N/m0.03 m2 

0.270 J

 2.5 kg  9.81m/s 2 0.03 m 
  0.736 J
U s  12 kA2  mgA

1
2
600 N/m0.03 m2
 2.5 kg 9.81m/s 2 0.03 m 
 1.01J
1
2
600 N/m0.03 m2
 0.270 J
*59 ••
Picture the Problem We can find the amplitude of the
motion by relating it to the maximum speed of the object.
Let the origin of our coordinate system be at y0, where y0
is the equilibrium position of the object and let Ug = 0 at
this location. Because Fnet = 0 at equilibrium, the
extension of the spring is then y0 = mg/k, and the potential
energy stored in the spring is U s  12 ky02 . A further extension
of the spring by an amount y increases Us to
2
2
2
2
2
1
1
1
1
1
2 k  y  y0   2 ky  kyy0  2 ky0  2 ky  mgy  2 ky0 . Consequently, if
we set
U = Ug + Us = 0, a further extension of the spring by y
increases Us by ½ky2 + mgy while decreasing Ug by mgy.
Therefore, if U = 0 at the equilibrium position, the change
in U is given by 12 k  y'2 , where y = y  y0.
(a) Relate the
maximum speed of the
object to the amplitude
of its motion:
Solve for A:
vmax  A
A
vmax

 vmax
Substitute numerical
values and evaluate A:
A  0.3 m/s 
(b) Express the energy
of the object at
maximum
displacement:
E  12 kA2
Substitute numerical
values and evaluate E:
E
(c) At maximum
displacement from
equilibrium:
U g  mgA
Substitute numerical
values and evaluate Ug:
(d) Express the
1
2
m
k
1.2 kg
 1.90 cm
300 N/m
300 N/m0.019 m2 


0.0542 J
U g  1.2 kg  9.81m/s 2 0.019 m
  0.224 J
U s  12 kA2  mgA
potential energy in the
spring when the object
is at its maximum
downward
displacement:
Substitute numerical
values and evaluate Us:
Us 
300 N/m0.019 m2
 1.2 kg 9.81m/s 2 0.019 m 
1
2
 0.278 J