Energetics of SHM - Mr. Mac's Physics...
69
ENERGETICS OF SHM
Transcript of Energetics of SHM - Mr. Mac's Physics...
ENERGETICS OF SHM
WORK DONE IN STRETCHING A SPRING
F
x
m
Work done ON the spring is positive; work BY spring is negative.
From Hooke’s law the force F is:
F (x) = kx
x1 x2
FTo stretch spring from
x1 to x2 , work is:
2 2
2 1½ ½Work kx kx 2 2
2 1½ ½Work kx kx
(Review module on work)
EXAMPLE 2(CONT.: THE MASS M IS NOW
STRETCHED A DISTANCE OF 8 CM AND HELD. WHAT
IS THE POTENTIAL ENERGY? (K = 196 N/M)
F8 cm
m
U = 0.627 JU = 0.627 J
The potential energy is equal to the work done in stretching the spring:
2 2
2 1½ ½Work kx kx 0
2 2½ ½(196 N/m)(0.08 m)U kx
CONSERVATION OF ENERGY
The total mechanical energy (U + K) of a vibrating system is constant; i.e., it is the same at any point in the oscillating path.
m
x = 0 x = +Ax = -A
xv
a
For any two points A and B, we may write:
½mvA2 + ½kxA
2 = ½mvB2 + ½kxB
2½mvA2 + ½kxA
2 = ½mvB2 + ½kxB
2
ENERGY OF A VIBRATING SYSTEM:
m
x = 0 x = +Ax = -A
x va
• At any other point: U + K = ½mv2 + ½kx2
U + K = ½kA2 x = A and v = 0.
• At points A and B, the velocity is zero and the
acceleration is a maximum. The total energy is:
A B
The reference circle
compares the circular
motion of an object with its
horizontal projection.
w 2f
THE REFERENCE CIRCLE
cos(2 )x A ft cos(2 )x A ft
cosx A t w
x = Horizontal displacement.
A = Amplitude (xmax).
= Reference angle.
EXAMPLE 7: AT WHAT TIME WILL THE 2-KG MASS
BE LOCATED 12 CM TO THE LEFT OF X = 0?
(A = 20 CM, F = 2.25 HZ)
m
x = 0 x = +0.2 m
x va
x = -0.2 m
t = 0.157 st = 0.157 s
cos(2 )x A ft
-0.12 m
10.12 mcos(2 ) ; (2 ) cos ( 0.60)
0.20 m
xft ft
A
2.214 rad2 2.214 rad;
2 (2.25 Hz)ft t
VELOCITY IN SHM
The velocity (v) of an oscillating body at any instant is the horizontal component of its tangential velocity (vT).
vT = wR = wA; w 2f
v = -vT sin ; = wt
v = -w A sin w t
v = -2f A sin 2f tv = -2f A sin 2f t
EXAMPLE 6: THE 2-KG MASS OF THE PREVIOUS
EXAMPLE IS DISPLACED INITIALLY AT X = 20 CM
AND RELEASED. WHAT IS THE VELOCITY 2.69 S
AFTER RELEASE? (RECALL THAT F = 2.25 HZ.)
m
x = 0 x = +0.2 m
x va
x = -0.2 m
v = -0.916 m/sv = -0.916 m/s
v = -2f A sin 2f tv = -2f A sin 2f t
2 (2.25 Hz)(0.2 m)sin 2 (2.25 Hz)(2.69 s)v
(Note: in rads) 2 (2.25 Hz)(0.2 m)(0.324)v
The minus sign means it is moving to the left.
The acceleration (a) of an oscillating body at any instant is the horizontal component of its centripetal acceleration (ac).
ACCELERATION REFERENCE CIRCLE
a = -ac cos = -ac cos(wt)
2 2 22; c c
v Ra a R
R R
ww
R = A
a = -w2A cos(wt)
2 24 cos(2 )a f A ft
2 24a f x
EXAMPLE 6 (CONT.): SUPPOSE THE 2-KG MASS
OF THE PREVIOUS PROBLEM IS DISPLACED 20
CM AND RELEASED (K = 400 N/M). WHAT IS THE
MAXIMUM ACCELERATION? (F = 2.25 HZ)
m
x = 0 x = +0.2 m
x va
x = -0.2 m
2 2 2 24 4 (2.25 Hz) ( 0.2 m)a f x
Acceleration is a maximum when x = A
a = 40 m/s2a = 40 m/s2
THE PERIOD AND FREQUENCY
GENERALIZED
For any body undergoing simple harmonic motion:
PERIOD AND FREQUENCY AS A FUNCTION
OF MASS AND SPRING CONSTANT.
For a vibrating body with an elastic restoring force:
Recall that F = ma = -kx:
1
2
kf
m
1
2
kf
m 2
mT
k 2
mT
k
The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units.
The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units.
EXAMPLE 6: THE FRICTIONLESS SYSTEM SHOWN
BELOW HAS A 2-KG MASS ATTACHED TO A
SPRING (K = 400 N/M). THE MASS IS DISPLACED
A DISTANCE OF 20 CM TO THE RIGHT AND
RELEASED.
WHAT IS THE FREQUENCY OF THE MOTION?
m
x = 0 x = +0.2 m
x va
x = -0.2 m
1 1 400 N/m
2 2 2 kg
kf
m f = 2.25 Hzf = 2.25 Hz
THE SIMPLE PENDULUM
The period of a simple
pendulum is given by:
mg
L
2L
Tg
For small angles .
1
2
gf
L
EXAMPLE 8. WHAT MUST BE THE LENGTH OF A
SIMPLE PENDULUM FOR A CLOCK WHICH HAS A
PERIOD OF TWO SECONDS (TICK-TOCK)?
2L
Tg
L
22 2
24 ; L =
4
L T gT
g
2 2
2
(2 s) (9.8 m/s )
4L
L = 0.993 m
THE TORSION PENDULUM
The period T of a torsion pendulum is given by:
Where k’ is a torsion constant that depends on the material from which the rod is made; I is the rotational inertia of the vibrating system.
Where k’ is a torsion constant that depends on the material from which the rod is made; I is the rotational inertia of the vibrating system.
2'
IT
k
EXAMPLE 9: A 160 G SOLID DISK IS ATTACHED TO THE
END OF A WIRE, THEN TWISTED AT 0.8 RAD AND
RELEASED. THE TORSION CONSTANT K’ IS 0.025 N
M/RAD. FIND THE PERIOD.
(Neglect the torsion in the wire)
For Disk: I = ½mR2
I = ½(0.16 kg)(0.12 m)2
= 0.00115 kg m2
20.00115 kg m2 2
' 0.025 N m/rad
IT
k T = 1.35 sT = 1.35 s
Note: Period is independent of angular displacement.
SUMMARY (CONT.)F
x
m
Hooke’s Law: In a spring, there is a restoringforce that is proportional to the displacement.Hooke’s Law: In a spring, there is a restoringforce that is proportional to the displacement.
The spring constant k is defined by:
Fk
x
Fk
x
F kx F kx
SUMMARY (SHM)
F ma kx F ma kx kx
am
kxa
m
m
x = 0 x = +Ax = -A
x va
½mvA2 + ½kxA
2 = ½mvB2 + ½kxB
2½mvA2 + ½kxA
2 = ½mvB2 + ½kxB
2
Conservation of Energy:
SUMMARY (SHM)
2 2kv A x
m 2 2k
v A xm
2 2 21 1 12 2 2mv kx kA 2 2 21 1 1
2 2 2mv kx kA
0
kv A
m0
kv A
m
cos(2 )x A ft cos(2 )x A ft
2 sin(2 )v fA ft 2 sin(2 )v fA ft
2 24a f x 2 24a f x
SUMMARY: PERIOD AND FREQUENCY
FOR VIBRATING SPRING.
m
x = 0 x = +Ax = -A
x va
2m
Tk
2m
Tk
1
2
kf
m
1
2
kf
m
SUMMARY: SIMPLE PENDULUM AND
TORSION PENDULUM
2L
Tg
1
2
gf
L
L
2'
IT
k
ENERGY AND SIMPLE HARMONIC
MOTION24
A spring also has potential
energy when the spring is
stretched or compressed,
which we refer to as elastic
potential energy. Because of
elastic potential energy, a
stretched or compressed
spring can do work on an
object that is attached to the
spring.
25
26
DEFINITION OF ELASTIC POTENTIAL ENERGY
The elastic potential energy PEelastic is the energy that a
spring has by virtue of being stretched or compressed.
For an ideal spring that has a spring constant k and is
stretched or compressed by an amount x relative to its
unstrained length, the elastic potential energy is
SI Unit of Elastic Potential Energy: joule (J)
2
2
1kxPEelastic
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28
EXAMPLE 7. AN OBJECT ON A
HORIZONTAL SPRING29
An object of mass m = 0.200 kg that is vibrating on a
horizontal frictionless table. The spring has a spring constant
k = 545 N/m. It is stretched initially to x0 = 4.50 cm and then
released from rest (see part A of the drawing). Determine the
final translational speed vf of the object when the final
displacement of the spring is (a) xf = 2.25 cm and (b) xf = 0 cm.
30
2
0
22
2
1
2
1
2
1kxkxmv ff
)( 22
0 ff xxm
kv
0EE f
2
00
2
0
2
0
222
2
1
2
1
2
1
2
1
2
1
2
1kxmghmmvkxmghImv ffff ww
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(a) Since x0 = 0.0450 m and xf = 0.0225 m,
(b) When x0 = 0.0450 m and xf = 0 m,
CONCEPTUAL EXAMPLE 8.
CHANGING THE MASS OF A SIMPLE
HARMONIC OSCILLATOR
32
A box of mass m attached to a
spring that has a force constant k.
The box rests on a horizontal,
frictionless surface. The spring is
initially stretched to x = A and then
released from rest. The box then
executes simple harmonic motion
that is characterized by a maximum
speed vmax, an amplitude A, and an
angular frequency w.
33
When the box is passing through the point where the spring
is unstrained (x = 0 m), a second box of the same mass m and
speed vmax is attached to it, as in part b of the drawing.
Discuss what happens to (a) the maximum speed, (b) the
amplitude, and (c) the angular frequency of the subsequent
simple harmonic motion.
(a) The maximum speed of the two-box system remains the
same as that of the one-box system.
(c) The angular frequency of the two-box system is smaller
than that of the one-box system by a factor of 2
(b) The amplitude of the two-box system is greater than
that of the one-box system by a factor of 2
Elastic Potential Energy A2
m
kw
2
max wAa
EXAMPLE 9.
A FALLING BALL ON A VERTICAL SPRING
34
A 0.20-kg ball is attached to a
vertical spring. The spring
constant of the spring is 28 N/m.
The ball, supported initially so that
the spring is neither stretched nor
compressed, is released from rest.
In the absence of air resistance,
how far does the ball fall before
being brought to a momentary stop
by the spring?
35
xf = –h0, h0 = 2mg/k.
CHECK YOUR UNDERSTANDING 3
36
A block is attached to the end of a horizontal ideal spring
and rests on a frictionless surface. The block is pulled so that
the spring stretches relative to its unstrained length. In each
of the following three cases, the spring is stretched initially
by the same amount, but the block is given different initial
speeds. Rank the amplitudes of the resulting simple
harmonic motion in decreasing order (largest first). (a) The
block is released from rest. (b) The block is given an initial
speed v0. (c) The block is given an initial speed v0/2.
(b), (c), (a)
THE PENDULUM
37
A simple pendulum consists of a
particle of mass m, attached to a
frictionless pivot P by a cable of
length L and negligible mass.
38
EXAMPLE 10. KEEPING TIME
Lgf /2
1
39
Determine the length of a simple pendulum that will swing
back and forth in simple harmonic motion with a period of
1.00 s.
EXAMPLE 11. PENDULUM MOTION
AND WALKING 40
When we walk, our legs alternately swing forward about the
hip joint as a pivot. In this motion the leg is acting
approximately as a physical pendulum. Treating the leg as a
uniform rod of length D = 0.80 m, find the time it takes for
the leg to swing forward.
The desired time is one-half of the period or 0.75 s.
CONCEPTUAL QUESTION 6
41
REASONING AND SOLUTION A block is attached to a
horizontal spring and slides back and forth in simple
harmonic motion on a frictionless horizontal surface. A
second identical block is suddenly attached to the first block
when the first block is at one extreme end of the oscillation
cycle.
a. Since the attachment is made at one extreme end of the
oscillation cycle, where the velocity is zero, the extreme
end of the oscillation cycle will remain at the same point;
in other words, the amplitude remains the same.
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b. The angular frequency of an object of mass m in simple
harmonic motion at the end of a spring of force constant k
is given by Equation 10.11: . Since the mass m
is doubled while the force constant k remains the same,
the angular frequency decreases by a factor of . The
vibrational frequency f is related to w by f = w/(2 ) ; the
vibrational frequency f will also decrease by a factor of
w k /m
2
2
c. The maximum speed of oscillation is given by Equation
10.8: . Since the amplitude, A, remains the
same and the angular frequency, w, decreases by a factor
of , the maximum speed of oscillation also decreases by a
factor of .
vmax Aw
2
2
CONCEPTUAL QUESTION 11
43
REASONING AND SOLUTION From Equations 10.5 and 10.11,
we can deduce that the period of the simple harmonic motion of
an ideal spring is given by , where m is the mass at
the end of the ideal spring and k is the spring constant. We can
deduce from Equations 10.5 and 10.16 that, for small angles, the
period, T, of a simple pendulum is given by where L
is the length of the pendulum.
T 2 m/ k
T 2 L /g
In principle, the motion of a simple pendulum and an object on an ideal
spring can both be used to provide the period of a clock. However, it is
clear from the expressions for the period given above that the period of
the mass-spring system depends only on the mass and the spring
constant, while the period of the pendulum depends on the acceleration
due to gravity. Therefore, a pendulum clock is likely to become more
inaccurate when it is carried to the top of a high mountain where the
value of g will be smaller than it is at sea level.
CONCEPTUAL QUESTION 12
44
REASONING AND SOLUTION We can deduce from
Equations 10.5 and 10.16 that, for small angles, the period, T,
of a simple pendulum is given by where L is the
length of the pendulum. This can be solved for the
acceleration due to gravity to yield:
T 2 L /g
g 42L /T
2
If you were held prisoner in a room and had only a watch and a pair
of shoes with shoelaces of known length, you could determine
whether this room is on earth or on the moon in the following way:
You could use one of the shoelaces and one of the shoes to make a
pendulum. You could then set the pendulum into oscillation and use
the watch to measure the period of the pendulum. The acceleration
due to gravity could then be calculated from the expression above. If
the value is close to 9.80 m/s2, then it can be concluded that the room
is on earth. If the value is close to 1.6 m/s2, then it can be concluded
that the room is on the moon.
PROBLEM 8
45
REASONING AND SOLUTION The figure at the right shows
the original situation before the spring is cut. The weight, W, of
the object stretches the string by an amount x.
kx
W
Applying F = kx to this situation gives W = kx (1)
46
The figure at the right shows the
situation after the spring is cut
into two segments of equal
length.
Let k' represent the spring
constant of each half of the
spring after it is cut. Now the
weight, W, of the object stretches
each segment by an amount x'.
Applying F = kx to this situation
gives
k'x'
W
k'x'
W = k'x' + k'x' = 2k'x' (2)
47
Combining Equations (1) and (2) yields
kx = 2k'x'
From Conceptual Example 2, we know that k' = 2k so
that
kx = 2(2k)x'
Solving for x' gives
x' x
4
0.160 m
4 0.040 m
PROBLEM 16
48
REASONING AND SOLUTION From Conceptual Example 2,
we know that when the spring is cut in half, the spring
constant for each half is twice as large as that of the original
spring. In this case, the spring is cut into four shorter springs.
Thus, each of the four shorter springs with 25 coils has a
spring constant of 4 420 N/m1680 N/m
The angular frequency of simple harmonic motion is given
by Equation 10.11:
w k
m
1680 N/m
46 kg 6.0 rad/s
PROBLEM 17
49
REASONING AND SOLUTION a. Since the object oscillates
between , the amplitude of the
motion is 0.08m
b. From the graph, the period is
T=4.0 s . Therefore, according
to Equation 10.4,
w 2
T
2
4.0 s 1.6 rad/s
50
d. At t=1.0 s, the graph shows that the spring has its
maximum displacement. At this location, the object is
momentarily at rest, so that its speed is v=0 m/s
e. The acceleration of the object at t=1.0 s is a maximum,
and its magnitude is
amax Aw2 (0.080 m)(1.6 rad/s)2 = 0.20 m/s2
k w 2m (1.6 rad/s)2 (0.80 kg) 2.0 N/m
c. Equation 10.11 relates the angular frequency to the spring
constant: . Solving for k we find w k /m
PROBLEM 21
51
REASONING The frequency of vibration of the spring is
related to the added mass m by Equations 10.6 and 10.11:
f 1
2
k
m
The spring constant can be determined from Equation 10.1
SOLUTION Since the spring stretches by 0.018 m when
a 2.8-kg object is suspended from its end, the spring
constant is, according to Equation 10.1,
52
2Applied 3(2.8 kg)(9.80 m/s )
1.52 10 N/m0.018 m
F mgk
x x
Solving Equation (1) for m, we find that the mass required
to make the spring vibrate at 3.0 Hz is
m k
42f2
1.52 103 N/m
42(3.0 Hz)
2 4.3 kg
m
kf
2
1
m
kf
2
2
4
1
224 f
km
PROBLEM 25
elasticEP
53
0.392m
0. 2m0. 2m
0. 2m
point of
release
2.0kg
PROBLEM 25
54
REASONING AND SOLUTION If we neglect air resistance, only
the conservative forces of the spring and gravity act on the ball.
Therefore, the principle of conservation of mechanical energy
applies
When the 2.00 kg object is hung on the end of the vertical spring,
it stretches the spring by an amount x, where
x F
kmg
k
(2.00 kg)(9.80 m/s2 )
50.0 N/m 0.392 m
This position represents the equilibrium position of the system
with the 2.00-kg object suspended from the spring. The object is
then pulled down another 0.200 m and released from rest
(v0=0 m/s).
55
At this point the spring is stretched by an amount
of .This point represents the zero
reference level ( m) for the gravitational potential energy.
0.392 m + 0.200m = 0.592 m
h 0
h = 0 m: The kinetic energy, the gravitational potential
energy, and the elastic potential energy at the point of release
are: 1 12 2
02 2KE (0 m/s) 0 Jmv m
gravityPE (0 m) 0 Jmgh mg
PEelastic1
2kx0
2
1
2(50.0 N/m)(0.592 m)
2 8.76 J
The total mechanical energy E0 at the point of release is the
sum of the three energies above: E0 8.76 J
56
h = 0.200 m: When the object has risen a distance of
above the release point, the spring is stretched by an amount
of . Since the total mechanical
energy is conserved, its value at this point is still .
The gravitational and elastic potential energies are:
h 0.200 m
0.592 m– 0.200 m= 0.392 m
E 8.76 J
PEgravity mgh (2.00 kg)(9.80 m/s2
)(0.200 m) 3.92 J
PEelastic1
2kx
2
1
2(50.0 N/m)(0.392 m)
2 3.84 J
KE E –PEgravity– PEelastic 8.76 J– 3.92 J– 3.84 J= 1.00 J
KE PEgravity PEelastic ESince
57
h = 0.400 m: When the object has risen a distance of
above the release point, the spring is stretched by an amount
of . At this point, the total
mechanical energy is still . The gravitational and
elastic potential energies are:
h 0.400 m
0.592 m– 0.400 m=0.192 mE 8.76 J
PEgravity mgh (2.00 kg)(9.80 m/s2
)(0.400 m) 7.84 J
PEelastic1
2kx
2
1
2(50.0 N/m)(0.192 m)
2 0.92 J
The kinetic energy is
JJJJPEPEEKE elasticgravity 092.084.776.8
58
The results are summarized in the table below
PEgravh KE E
0.000 m 0.00 J 0.00 J 8.76 J 8.76 J
0.200 m 1.00 J 3.92 J 3.84 J 8.76 J
0.400 m 0.00 J 7.84 J 0.92 J 8.76 J
PEelastic
PROBLEM 32
59
m
f = 3.0 HZ
m/2
A = 5.08*10-2 m
Max speed at halfway of the amplitude.
m m/2
k
k
60
REASONING AND SOLUTION
a. Now look at conservation of energy before and after the split
Before split (1/2) mvmax2 = (1/2) kA2
Solving for the amplitude A gives
A vmax
m
k
After split If new amplitude is A’
(1/2) (m/2)v'2 = (1/2) (m/2)(vmax)2 = (1/2) kA'2
61
Solving for the amplitude A' gives
A vmax
m
2k
Therefore, we find that
A' = A/ = (5.08 * 10–2 m)/ = 3.59*10-2m2 2
Similarly, for the frequency, we can show that
f' = f = (3.00 Hz) = 4.24 Hz2 2
m
kw
m
kf 2
m
kf
2
1
62
b. If the block splits at one of the extreme positions,
the amplitude of the SHM would not change, so it
would remain as
The frequency would be
f' = f = (3.00 Hz) = 4.24 Hz2 2
5.08*10-2 m
PROBLEM 38
ond
mk
gx sec015.0
63
.x
Object is resting on the spring.
F = kx = mg
.. t = 0.25 s
f = 12
km
x = g
km
= 0.0155 m.
64
REASONING AND SOLUTION
Using f = 1/T = 1/(0.250 s) = 4.00 Hz and also
we can find the ratio k/m = 4 2f2 = 632 N/(kgm)
With the object resting on the spring, F = kx = mg so that,
65
When the mass leaves the spring, potential energy of the
spring has been converted to gravitational energy, i.e.,
(1/2) kx'2 = mgh
Where x' = 0.0500 m + 0.0155 m = 0.0655 m
Solving for h we get
h k
m
x' 2
2g
632 N/(kgm)
(0.0655 m)2
2(9.80 m/s2
)
0.138 m
if h is the height, it can reach
PROBLEM 41
66
REASONING AND SOLUTION Recall that the relationship
between frequency f and period T is . Then, according
to Equations 10.6 and 10.16, the period of the simple
pendulum is given by
f 1/T
T 2 L
g
where L is the length of the pendulum. Solving for g and
noting that the period is T = (280 s)/100 = 2.8 s, we obtain
g 42L
T2
42(1.2 m)
(2.8 s)2 6.0 m/s
2
PROBLEM 68
w
w
67
REASONING The force F that the spring exerts on the block
just before it is released is equal to –kx, according to Equation
10.2. Here k is the spring constant and x is the displacement of
the spring from its equilibrium position. Once the block has
been released, it oscillates back and forth with an angular
frequency given by Equation 10.11 as , where m is
the mass of the block. The maximum speed that the block
attains during the oscillatory motion is vmax = A (Equation
10.8). The magnitude of the maximum acceleration that the
block attains is amax = A 2 (Equation 10.10).
/k mw
68
SOLUTION
a. The force F exerted on the block by the spring is
82.0 N/m 0.120 m 9.84 NF kx
b. The angular frequency of the resulting oscillatory
motion is
82.0 N /m10.5 rad /s
0.750 kg
k
mw
w
69
d. The magnitude amax of the maximum acceleration is
22 2
max 0.120 m 10.5 rad /s 13.2 m/sa Aw
max 0.120 m 10.5 rad /s 1.26 m/sv Aw
c. The maximum speed vmax is the product of the
amplitude and the angular frequency:
