The document discusses a 0.5 kg object attached to a spring with a constant of 20 N/m oscillating on a frictionless surface. (a) Given an amplitude of 3 cm, the total energy and maximum speed are calculated. (b) The velocity at a displacement of 2 cm is found. (c) The kinetic and potential energies are computed at a given displacement.

1. The Period of a Mass on a Spring
Example: A mass of m = 0.12 kg attached to a spring
oscillates with an amplitude of 0.075 m and a maximum
speed of 0.524 m/s. Find (a) the spring constant, (b) the
period of motion.
€
v = −Aω sin(ωt) = −Aω sin(
2π
T
t)
vmax occurs for sin(ωt) =1⇒ ωt =
π
2
The maximum velocity occurs when mass passes the equilibrium position.
vmax = Aω ⇒ ω =
vmax
A
=
0.524m /s
0.075m
= 7s-1
(a) For a mass on a string: ω =
k
m
⇒ k = ω2
m = (7s-1
)2
(0.12kg) = 5.86 kg s-2
(b) T =
2π
ω
=
2π
7s-1 = 0.9s

2. The
Period
of
a
Mass
on
a
Spring
Example:
A
mass
of
m
=
0.26
kg
is
abached
to
a
ver:cal
spring.
When
in
mo:on
the
period
is
T
=
1.12s.
(a)
How
much
does
the
mass
stretch
the
spring
when
it
is
in
its
equilibrium
posi:on?
(b)
Suppose
this
experiment
was
performed
on
a
planet
with
twice
the
accelera:on
of
gravity
on
Earth.
By
what
factors
do
the
period
and
equilibrium
posi:on
change?
€
(a) When in equilibrium on Earth:(a) mgE = kyE ⇒ yE =
mgE
k
(b) TP = 2π
m
k
= TE , m and k do not depend on gravity
and therefore the period of oscillation will not change on
planet X. The equilibrium position on planet X will be:
yP =
m2gE
k
= 2yE

3. Energy
Conserva:on
in
Oscillatory
Mo:on
Stop
the
Block:
A
m
=
0.98kg
block
with
v=1.32
m/s
encounters
a
spring
with
k=245
N/m.
(a)
How
far
is
the
spring
compressed
before
coming
to
a
rest?
(b)
How
long
is
the
block
in
contact
with
the
spring
before
it
comes
to
rest?
€
(a) Conservation of Energy:Ei = Ef ⇒
1
2
mv2
=
1
2
kA2
⇒
A2
=
mv2
k
⇒ A = v
m
k
⇒ A =1.32m/s
0.98kg
245N/m
= 0.083m
(b) The block is in contact for a quarter period: t = T/4
t =
T
4
=
2π
4
m
k
=
2π
4
A
v
=
π
2
0.083m
1.32m/s
= 0.1s

4. Energy
Conserva:on
in
Oscillatory
Mo:on
Bullet-­‐Block
Collision:
A
bullet
of
mass
m
embeds
itself
in
a
block
of
mass
M.
If
the
ini:al
speed
of
the
bullet
is
v0
,
find
(a)
the
maximum
compression
of
the
spring,
(b)
the
:me
for
the
bullet-­‐block
to
come
to
rest.
€
(a) Conservation of momentum: Pi = Pf ⇒ mv0 = (m +M)vf ⇒ vf =
mv0
(m +M)
Conservsation of Mechanical Energy: Ei = Ef ⇒
1
2
(m+ M)vf
2
=
1
2
kA2
⇒
1
2
(m+ M)
mv0
(m +M)
⎡
⎣
⎢
⎤
⎦
⎥
2
=
1
2
kA2
⇒
mv0
( )
2
(m+ M)
= kA2
⇒ A =
mv0
( )
k(m+ M)
(b) The block is in contact t =
T
4
=
2π
4
m +M
k

5. Simple Harmonic Motion on a Frictionless Surface: A 0.350-kg object
attached to a spring of force constant 1.30 × 102 N/m is free to move on a
frictionless horizontal surface, as in Active Figure 13.1. If the object is released
from rest at x = 0.100 m, find the force on it and its acceleration at x = 0.100 m, x
= 0.050 0 m, x = 0 m, x = 20.050 m, and x = 20.100 m.
m = 0.350kg, k =1.3 ×102
N/m, A = 0.1m
ω2
=
k
m
=
1.3 ×102
N/m
0.350kg
= 371.4
N
m kg
F(x) = −kx
a(x) = −Aω2
cos ωt
( ) = −ω2
x

6. Mass
on
a
Ver7cal
Spring:
A
spring
is
hung
ver:cally
(Fig.
13.2a),
and
an
object
of
mass
m
abached
to
the
lower
end
is
then
slowly
lowered
a
distance
d
to
the
equilibrium
point
(Fig.
13.2b).
(a)
Find
the
value
of
the
spring
constant
if
the
spring
is
displaced
by
2.0
cm
and
the
mass
is
0.55
kg.
(b)
If
a
second
iden:cal
spring
is
abached
to
the
object
in
parallel
with
the
first
spring
(Fig.
13.2d),
where
is
the
new
equilibrium
point
of
the
system?
(c)
What
is
the
effec:ve
spring
constant
of
the
two
springs
ac:ng
as
one?
€
(a) mg = kd ⇒ k =
mg
d
=
0.55kg
( ) 9.80m/s2
( )
0.02m
= 269.5kg/s2
(b) mg = 2kd1 ⇒ d1 =
mg
2k
=
0.55kg
( ) 9.80m/s2
( )
2 269.5kg/s2
( )
= 0.01m
(c) mg = keff d1 ⇒ keff =
mg
d1
= 539kg/s2

7. Stop
That
Car!:
A
13
000-­‐N
car
starts
at
rest
and
rolls
down
a
hill
from
a
height
of
10.0
m
(Fig.
13.6).
It
then
moves
across
a
level
surface
and
collides
with
a
light
spring-­‐
loaded
guard-­‐rail.
(a)
Neglec:ng
any
losses
due
to
fric:on,
and
ignoring
the
rota:onal
kine:c
energy
of
the
wheels,
find
the
maximum
distance
the
spring
is
compressed.
Assume
a
spring
constant
of
1.0
×
106
N/m.
(b)
Calculate
the
maximum
accelera:on
of
the
car
amer
contact
with
the
spring,
assuming
no
fric:onal
losses.
(c)
If
the
spring
is
compressed
by
only
0.30
m,
find
the
change
in
the
mechanical
energy
due
to
fric:on.
€
(a) Conservation of energy: PE →KE →PE
mgh =
1
2
mv2
=
1
2
kx2
⇒ x2
=
2mgh
k
⇒ x =
2mgh
k
=
2 13000
( ) 10m
( )
1.0 ×106
N/m
= 0.51m
(b) a = -Aω2
cos ωt
( ) ⇒ amax = Aω2
(1)
ω =
2π
T
=
k
m
=
1.0 ×106
N/m
13000 N
( )/ 9.8 m/s2
( )
= 27.5 Hz (2)
(1)^(2) ⇒ amax = 0.51m
( ) 27.5 Hz
( )2
= 385.7m/s2
(c) Wnc = ΔE = E f - Ei =
1
2
k(0.3m)2
− (mgh) ⇒
0.5
( ) 1.0 ×106
N/m
( )(0.3m)2
− 13000 N
( ) 10m
( ) = −8.5 ×105
J

8. The
Object–Spring
System
Revisited:
A
0.500-­‐kg
object
connected
to
a
light
spring
with
a
spring
constant
of
20.0
N/m
oscillates
on
a
fric:on-­‐less
horizontal
surface.
(a)
Calculate
the
total
energy
of
the
system
and
the
maximum
speed
of
the
object
if
the
amplitude
of
the
mo:on
is
3.00
cm.
(b)
What
is
the
velocity
of
the
object
when
the
displacement
is
2.00
cm?
(c)
Compute
the
kine:c
and
poten:al
energies
of
the
system
when
the
displacement
is
2.00
cm.
€
(a) Etot =
1
2
kA2
= 0.5
( ) 20 N/m
( ) 0.03m
( )2
= 0.009J
(b)
Etot = 0.009J =
1
2
mv2
+
1
2
k(0.02m)2
⇒ v = ±0.141m/s
(c) KE(0.02m) =
1
2
mv2
= 0.5 0.5kg
( ) 0.14m/s
( )2
= 0.005J
PE = 0.009J - 0.005J = 0.004J or
PE =
1
2
kx2
= 0.5 20 N/m
( ) 0.02m
( )2
= 0.004J

9. That
Car
Needs
Shock
Absorbers!:
A
1.30
×
103-­‐kg
car
is
constructed
on
a
frame
supported
by
four
springs.
Each
spring
has
a
spring
constant
of
2.00
×
104
N/m.
If
two
people
riding
in
the
car
have
a
combined
mass
of
1.60
×
102
kg,
find
the
frequency
of
vibra:on
of
the
car
when
it
is
driven
over
a
pothole
in
the
road.
Find
also
the
period
and
the
angular
frequency.
Assume
the
weight
is
evenly
distributed.
€
Effective spring constant :
mtotalg = 4kx ⇒ keff = 4k
ω =
4k
m
=
4 2 ×104
N/m
( )
1.46 ×103
kg
= 7.4Hz
ω =
2π
T
= 2πf ⇒ f =
ω
2π
=
7.4Hz
2π
=1.18Hz
T =
2π
ω
=
2π
7.4Hz
= 0.85s

10. The
Vibra7ng
Object–Spring
System:
(a)
Find
the
amplitude,
frequency,
and
period
of
mo:on
for
an
object
vibra:ng
at
the
end
of
a
horizontal
spring
if
the
equa:on
for
its
posi:on
as
a
func:on
of
:me
is
(b)
Find
the
maximum
magnitude
of
the
velocity
and
accelera:on.
(c)
What
are
the
posi:on,
velocity,
and
accelera:on
of
the
object
amer
1.00
s
has
elapsed?
€
x = 0.250 m
( )cos
π
8.00
t
⎛
⎝
⎜
⎞
⎠
⎟

11. The
Vibra7ng
Object–Spring
System:
(a)
Find
the
amplitude,
frequency,
and
period
of
mo:on
for
an
object
vibra:ng
at
the
end
of
a
horizontal
spring
if
the
equa:on
for
its
posi:on
as
a
func:on
of
:me
is
(b)
Find
the
maximum
magnitude
of
the
velocity
and
accelera:on.
(c)
What
are
the
posi:on,
velocity,
and
accelera:on
of
the
object
amer
1.00
s
has
elapsed?
€
x = 0.250 m
( )cos
π
8.00
t
⎛
⎝
⎜
⎞
⎠
⎟
€
(a) ω =
π
8
Hz A = 0.250m f =
ω
2π
=
π
8
Hz
2π
=
1
16
Hz T =
2π
ω
=
2π
π
8
Hz
=16s
(b)
vmax = - Aω = − 0.250m
( )
π
8
Hz
⎛
⎝
⎜
⎞
⎠
⎟ = 0.098 m/s
amax = -Aω2
= − 0.250m
( )
π
8
Hz
⎛
⎝
⎜
⎞
⎠
⎟
2
= 0.039 m/s2
(c) x(1s) = (0.25m)cos
π
8
v(1s) = −(0.25m)
π
8
Hz
⎛
⎝
⎜
⎞
⎠
⎟sin
π
8
a(1s) = −(0.25m)
π
8
Hz
⎛
⎝
⎜
⎞
⎠
⎟
2
cos
π
8

12. €
A Traveling Wave: A wave traveling in the positive x-direction is pictured in Figure
13.27a. Find the amplitude, wavelength, speed, and period of the wave if it has a
frequency of 8.00 Hz. In Figure 13.27a, Δx = 40.0 cm and Δy = 15.0 cm.
frequency: f = 8.00Hz
Amplitude : A = Δy =15.0 cm
Wavelength :λ = Δx = 40.0 cm
Speed : v = λf = (0.40 m)(8.00Hz) =3.2m/s
Period:P =1/f =1/8.00Hz =0.125s

13. Sound and Light: A wave has a wavelength of 3.00 m. Calculate the
frequency of the wave if it is (a) a sound wave and (b) a light wave. Take
the speed of sound as 343 m/s and the speed of light as 3.00 × 108 m/s.
€
λ = 3.00m
v = λf ⇒ f = v /λ
(a) Sound wave : f = v /λ =
343m/s
3.00m
=114.33Hz
(b) Light wave : f = c /λ =
3 ×108
m/s
3.00m
=100MHz

14. A Pulse Traveling on a String: A uniform string has a mass M of 0.030 kg
and a length L of 6.00 m. Tension is maintained in the string by suspending
a block of mass m = 2.00 kg from one end (Fig. 13.28). (a) Find the speed of
a transverse wave pulse on this string. (b) Find the time it takes the pulse to
travel from the wall to the pulley. Neglect the mass of the hanging part of the
string.
€
The velocity of a wave on a string is: v =
F
µ
,
where F is the tension in the string and µ is the mass per unit length of the string.
Using Newton's 2nd Law : F = mg = 2.00kg
( ) 9.8m/s2
( ) =19.6 N
Mass per unit length of the string: µ = 0.03kg/6.00m = 0.005 kg/m
v =
19.6 N
0.005 kg/m
= 62.6m/s
t = 5m
( )/ 10.44m/s
( ) = 0.08s

15. Sound
Waves
Example
:
You
drop
a
stone
in
a
7.35
m
deep
well.
How
long
before
you
hear
the
splash?

16. Sound
Waves
Example
:
You
drop
a
stone
in
a
7.35
m
deep
well.
How
long
before
you
hear
the
splash?
€
The equation of motion of the stone:
d =
1
2
gt1
2
⇒ t1
2
=
2d
g
⇒ t1 =
2(7.35m)
9.81 m/s2 =1.22 s
Time for sound from splash to reach you:
t2 = d/vs = (3.75 m)/(343 m/s) = 0.011 s
The total time is:
t = t1 +t2 =1.23 s