Download Equations apply to all oscillators

Oscillations
x
Oscillations are caused by a vibrating object.
m
Known as: Simple Harmonic Motion, SHM .
To oscillate an object must be displaced a
distance x , and the following equations apply:
F = kx
m
1 2
U = kx
2
m
Equations apply to all oscillators (not just springs)
θ
Maximum displacement, xmax , is known as amplitude, A .
A = xmax
There are four amplitudes in one cycle.
L
x
x
Oscillations are another type of cyclic motion.
Period
Time of one cycle
t
T=
# cycles
1
T=
f
For elastic devices, such as springs, there is a specific period
m
Ts = 2p
k
The period of an oscillating spring depends on the mass
attached to the spring, m , and the spring constant, k .
How does doubling the mass affect the period of an oscillating spring?
( 2 )T = 2p
m
Ts = 2p
k
The new period is
( 2) m
s
k
2T
What mass is needed to double the period of an oscillating spring?
Ts = 2p
m
k
( 2)T = 2p
The mass must be quadrupled
s
4m
(2 ) m
2
k
An object completes 20 cycles in 5.0 seconds. Determine
the period.
t
T=
# cycles
5.0s )
(
T=
= 0.25s
( 20)
A 500 g mass is suspended from a spring and stretches the
spring 20 cm to its equilibrium position. The spring is
then stretched an additional 10 cm and is released from
rest. Determine the frequency of the spring’s oscillations.
The key to spring problems is determining the spring constant, k .
Equilibrium, where forces are equal and opposite, is the most used method.
Fs = Fg
kx = mg
m
Ts = 2p
k
mg ( 0.500 ) (10 )
k=
=
= 25N m
x
( 0.20)
T = 2p
0.500
= 0.889s
25
f =
1
1
=
= 1.13 Hz
T 0.889
Spring problems are usually one of two types. They are either a force problem or an
energy problem. The following example demonstrates two common spring problems.
Energy
Force
m
x1
m
x2
x0 = 0
m
m
m
x2
A 10 kg mass is hung from a spring and lowered to
equilibrium displacing the mass 40 cm.
(A) Draw the FBD for the mass
Force
Fs
m
Fs
x1
m
Fg
Fg
A 10 kg mass is hung from a spring and lowered to
equilibrium displacing the mass 40 cm.
(B) Determine the spring constant k .
Force
Fs = Fg
Fs
m
Fs
x1
Fg
k x1 = mg
k ( 0.40 ) = (10) (10)
k = 250 N m
m
Fg
A 10 kg mass is hung from a spring and lowered to
equilibrium displacing the mass 40 cm.
Energy
Force
As the force problem
changes to an energy
problem x1 used in
force no longer matters.
For energy the
equilibrium
position becomes
Fs
x1
m
m
Fg
x0 = 0
The spring is stretched an additional 20 cm, beyond the
equilibrium position, and is released.
(C) Determine the total energy stored in the spring/mass system.
1
U s = k x2 2
2
Energy
2
1
U s = ( 250 ) ( 0.20 )
2
U s = 5.0J
x0 = 0
m
m
x2
The spring is stretched an additional 20 cm, beyond the
equilibrium position, and is released.
The released mass will
oscillate 20 cm above and
below equilibrium.
Energy
m
x2 = xmax
The displacement x2 is
the farthest (maximum)
displacement from the
equilibrium position.
x2 = xmax
x0 = 0
m
m
m
x2 = xmax
This is also the amplitude
of the spring’s oscillation.
The spring is stretched an additional 20 cm, beyond the
equilibrium position, and is released.
(D) Determine the speed when the mass is 10 cm from equilibrium.
Use conservation of energy.
U 0 + K0 = U + K
Set maximum displacement as the initial position,
since this is where everything is known.
m
0.20 m
The displacement was given, xmax = ±0.20 m
The velocity is always zero, v0 = 0 .
0.00 m
m
0.20 m
2
2
1
1
1 2 1 2
k ( 0.20 ) + m( 0) = k x + mv
2
2
2
2
The spring is stretched an additional 20 cm, beyond the
equilibrium position, and is released.
(D) Determine the speed when the mass is 10 cm from equilibrium.
2
2
1
1
1 2 1 2
k ( 0.20 ) + m( 0) = k x + mv
2
2
2
2
The final position is where information is needed.
The speed, v , at 10 cm is requested.
m
0.20 m
0.00 m
m
0.01 m
m
0.20 m
Set final displacement as 10 cm, x = 0.10 m .
2
2
1
1
1
2
250
0.20
+
0
=
250
0.10
+
10
v
( )( )
( )( ) 2 ( )
2
2
v = 0.866 m s
The spring is stretched an additional 20 cm, beyond the
equilibrium position, and is released.
(E) Determine the maximum speed during the oscillation.
As before, set maximum displacement as the
initial position, xmax = 0.20 m , and v0 = 0 .
2
2
1
1
1 2 1 2
k ( 0.20 ) + m( 0) = k x + mv
2
2
2
2
m
m
m
0.20 m
0.00 m
0.20 m
Maximum speed, vmax , occurs when
displacement and potential energy are zero, x = 0
.
2
2
1
1
1
2
250
0.20
+
0
=
250
0
+
10
v
( )( )
( )( ) 2 ( ) max
2
2
vmax = 1.0m s
The spring is stretched an additional 20 cm, beyond the
equilibrium position, and is released.
(F) Determine the period of oscillation.
Period is the time of one cycle.
Use the formula that is specific for a spring.
m
Ts = 2p
k
m
0.20 m
0.00 m
m
0.20 m
The mass was given at the beginning of the
problem, m = 10 kg .
Ts = 2p
(10)
( 250)
Ts = 1.26s
Since period is a time, it is measured in seconds.
Clearly height was changing during the oscillation.
Why was the energy associated with height, mgh ,
not mentioned in parts “f “ and “g” ?
During a vertical oscillation, the up/down motion
causes the gravitational potential energy to cancel.
m
0.20 m
When a spring oscillates vertically, ignore mgh .
However, if a spring moves in one direction,
0.00 m
m
0.20 m
without oscillating (such as from +xmax to −xmax ),
then the potential energy of gravity, mgh , is
included in the conservation of energy formula.
m
+xmax
x0 = 0
m
−xmax
x Us v
max max 0
0
K
åF
0 max max
0 max max 0
max max 0
a
0
0 max max
L
TP = 2p
g
For a pendulum with a string length L .
A pendulum is brought to a distant planet. The string of
the pendulum is adjusted until the frequency of the
pendulum swing is 1 cycle per second. The length of the
string is 2.5 m. Determine the acceleration of gravity.
L
TP = 2p
g
1
1
T= =
= 1.0s
f 1.0
(1.0) = 2p
( 2.5)
g
To solve a variable under a square root, square everything.
æ
2
(1.0) = 22 p 2 çç
è
( 2.5) ö÷
g ÷ø
2
(1.0) = 4p
2
( 2.5)
g
g = 98.7 m s2
Draw horizontal lines thru points A and B .
ϕ
θ
L
This creates a right triangle.
The vertical side, adjacent to the angle θ , has
a length of L cos θ
L cos θ
L
A ϕ
x
h
B
The sum of L cos θ and h is length L .
Lcosq + h = L
h = L - Lcosq
What if this angle ϕ is given instead of θ ?
What is displacement x is given?
h = L - Lsinq
h= L- L - x
2
2
A mass of 5.0 kg hangs from a 1.0 m string. The mass is
displaced from point B to point A . At point A the angle
that the string makes with the vertical is θ = 37o.
(A) Determine the restoring force acting on the
pendulum bob when it is at point A .
θ
Frestoring = Fg sinq
L
Frestoring = (5.0) (10) sin37 o = 30 N
(B) Determine the instantaneous
acceleration of the pendulum
bob at point A .
SF = Frestoring
A
a
Fg
SF = Fg sinq
(5.0) a = (5.0)(10) sin37
B
o
a = 6.0 m s2
The restoring force, sum of forces, and acceleration are all tangential at point A .
A mass of 5.0 kg hangs from a 1.0 m string. The mass is
displaced from point B to point A . At point A the angle
that the string makes with the vertical is θ = 37o.
(C) Determine the height between points A and B .
h = L - Lcosq
θ
h = (1.0 ) - (1.0 ) cos37 = 0.20 m
L
o
(D) Determine the energy stored
in the pendulum at point A .
U g = mgh
U g = (5.0 ) (10) ( 0.20 ) = 10J
A
h
B
A mass of 5.0 kg hangs from a 1.0 m string. The mass is
displaced from point B to point A . At point A the angle
that the string makes with the vertical is θ = 37o.
The pendulum is released from rest at point A .
(E) Determine the speed when height of 0.10 m .
U 0 + K0 = U + K
1
1
2
mgh0 + mv0 = mgh + mv 2
2
2
At maximum displacement
(position A) height, h = 0.20 m ,
and speed, v = 0 , are known.
θ
L
A
0.10 m
0.20 m
Substitute 0.10 m for the final
height to find the matching v .
1 2
1 2
(10)( 0.20) + 2 ( 0) = (10)(0.10) + 2 v
B
v = 1.41m s
A mass of 5.0 kg hangs from a 1.0 m string. The mass is
displaced from point B to point A . At point A the angle
that the string makes with the vertical is θ = 37o.
(F) Determine the maximum speed reached
by the pendulum.
θ
U 0 + K0 = U + K
1
1
2
mgh0 + mv0 = mgh + mv 2
2
2
At maximum displacement
(position A) height, h = 0.20 m,
and speed, v = 0 , are known.
L
A
0.20 m
Maximum speed, vmax , occurs at
equilibrium (position B), h = 0 .
1 2
1
(10)(0.20) + 2 (0) = (10)(0) + 2 vmax 2
B
vmax
vmax = 2.00m s
A mass of 5.0 kg hangs from a 1.0 m string. The mass is
displaced from point B to point A . At point A the angle
that the string makes with the vertical is θ = 37o.
(G) Determine the period of oscillation.
θ
Period is the time of one cycle.
Use the period of a pendulum equation.
L
Tp = 2p
g
TP = 2p
(1.0)
(10)
L
A
Tp = 1.99s
Since period is a time, it is measured in seconds.
B
(H) Complete the chart listing when quantities are maximum or zero, and
then sketch the corresponding potential, kinetic, and total energy graphs.
L
θ
U
−x
+x
–x
x=0
+x
x=0
K
x&h
max
0
max
US & Ug
max
0
max
v
0
max
0
K
0
max
0
ΣF
max
0
max
a
max
0
max
–x
x=0
+x
Total energy = ΣE = U + K
ΣE
–x
x=0
+x
These are the graphs of any object
in simple harmonic motion.
This includes all types of springs
U
–x
x=0
+x
Coil, leaf, etc.
K
And all types of pendulums
Traditional, physical, torsion, etc.
–x
Or, any other object experiencing
simple harmonic motion.
x=0
+x
ΣE
–x
x=0
+x
The total energy of an oscillator is the sum of potential and kinetic
energy.
E =U + K
1 2 1 2
E = kx + mv
2
2
If energy is doubled, how does this affect amplitude of an oscillator?
At amplitude: x = A and the kinetic energy is zero
1 2
E = kA
2
( )
1
2E = k ?A
2
2
1
2E = k
2
( 2 A)
2
Doubling energy would cause the amplitude to increase by the square
root of two.
The total energy of an oscillator is the sum of potential and kinetic
energy.
E =U + K
1 2 1 2
E = kx + mv
2
2
If energy is doubled, how does this affect the maximum speed of an
oscillator?
At maximum speed the elastic potential energy is zero.
1
E = mvmax 2
2
(
1
2E = m ?vmax
2
)
2
1
2E = m
2
(
2vmax
Doubling energy would cause the maximum speed to increase by the
square root of two.
)
2