Simple Harmonic Motion: Abdalla, Aymen Spring 2002 SC.441.L.H, Sec (8773) Instructor: Dr. Roman Kezerashvili May – 17 - 2002
Download ReportTranscript Simple Harmonic Motion: Abdalla, Aymen Spring 2002 SC.441.L.H, Sec (8773) Instructor: Dr. Roman Kezerashvili May – 17 - 2002
Slide 1
Simple Harmonic
Motion:
Abdalla, Aymen
Spring 2002
SC.441.L.H, Sec (8773) Instructor: Dr. Roman
Kezerashvili
May – 17 - 2002
Slide 2
Introduction:
Periodic is any motion that
repeats it self in any equal
interval of time. A vibrating
spring and a simple
pendulum exhibit periodic
motion. The simple
harmonic motion is a
special type of periodic
motion.
Slide 3
Objectives
To study the simple harmonic
motion by investigating the
period of oscillation of a
spring, and to determine the
constant of the spring for one
spring and two springs in
series.
Slide 4
Equipments:
1.
2.
3.
4.
5.
Two springs.
Triple-beam balance.
Photo gate accessory.
Science workshop Interface box.
Set of masses.
Slide 5
Theory:
Consider a mass m, attached
to a spring. When the spring
in a stretched position, a
force F acts on the mass, and
x is the distance the mass
moves from its equilibrium.
This force tends to restore
the mass to its original
position and it is called the
restoring force. Also it is
opposite to the displacement
x and from hooks law.
Slide 6
F = -kx
(1)
K is the force constant of the spring.
From Newton’s second law :
a = F/m
Combine (1) and (2)
a = -kx/m or a= d^2x/dt^2 = -Kx/m (3)
(2)
Slide 7
But a varies with x, that:
^2 = k/m
or
= (k/m)
(4)
From equation (3)
A = -^2 x or d^2x/dt^2 = -^2x
(5)
And from the differential equation :
x = Acos(t + )
(6)
Slide 8
And from the differential equation
:
x = Acos(t + )
(6)
A is the maximum displacement
and it is called amplitude, = /2
is the frequency of vibration, is
the angular frequency. And (t +
) is called the phase of the simple
harmonic motion, is called the
phase constant.
Slide 9
For x from A to –A and back to A is
called a cycle ant T is the time for one
complete oscillation (cycle). That:
cos(t + + 2) = cos(t + )
So that:
(t + T) + = t + + 2 or t = 2(8)
So:
T = 2/
(7)
(9)
Slide 10
Sub from (4) to (9):
T= 2 m/k
(10)
This is true when the mass of the spring
ms is much less than the mass m, which is
suspended from the spring. And for a
spring of finite mass ms is:
T = 2 (m + ms/3)/k
(11)
From (11) determine k as:
k = 4^2(m + ms/3)/T^2
(12)
Slide 11
If we have two springs in series with the
force constant k1, k2 and x is the sum of
the displacement of each spring, that is:
x = x1 + x2
From Hook’s law:
x = F1/k1,
(13)
x2 = F2/k2, and x =F/k
Sub these values in (14):
F/k = F1/k1 + F2/k2
(15)
(14)
Slide 12
Because the springs in series
F = F1 = F2, that:
1/k = 1/k1 + 1/k2
or:
So that T is:
T = 2m(1/k1 + 1/k2)
(16)
k = k1k2/(k1+k2) (17)
(18)
Slide 13
Procedure:
In this experiment we measure the period of
oscillation, T for a spring with mass ms and mass
m for the object. Then use (12) to determine k,
and repeat the same for a series of two springs.
Use the science workshop to measure Frequency
and Number of cycles.
1. Measure the mass of the spring.
2. Suspend the mass of the object larger than the
mass of the spring.
3. Start record T period and frequency.
4. Increase the mass m and repeat step 4 for total
5-7 trails.
5. Connect two springs in series and suspend
mass m larger than the mass of the two springs,
and repeat steps 5 & 5.
Slide 14
Mass of the first single spring = 0.1805
Total suspended mass
m,kg
Period T,s
Frequency f, Square
Hz of period
T^2, s^2
K from
equation
(12), N/m
.2
1.075
.931
1.16
8.86
.25
1.19
.841
1.42
8.64
.3
1.28
.781
1.64
8.67
.35
1.363
.734
1.86
8.7
.4
1.44
.694
2.07
8.8
Slide 15
m versus T^2
m,
k
g
1.16
0.2
1.42
0.25
1.64
0.3
1.86
0.35
0.5
y = 0.2209x - 0.06
0.4
m, Kg
T^2,
s^2
0.3
0.2
0.1
0
0
2.07
0.5
0.4
1
1.5
2
T^2, s^2
Mean value for k, N/m
30.36
K from the slope of the graph
30.32
% difference
.13%
2.5
Slide 16
Mass of the second single spring = .035kg
Total suspended
mass m,kg
Period
T,s
Frequency Square
f, Hz of period
T^2, s^2
K from
equation
(12), N/m
.2
.524
1.91
.28
29.85
.25
.583
1.715
.34
30.5
.3
.638
1.567
.4
30.79
.35
.687
1.456
.47
30.5
.4
.734
1.362
.54
30.16
Slide 17
m versus T^2
m,
kg
0.28
0.2
0.34
0.25
0.4
0.47
0.3
0.35
0.5
0.4
m, kg
T^2,
s
^
2
y = 0.768x - 0.0118
0.3
0.2
0.1
0
0
0.2
0.4
T^2, s^2
0.54
0.4
Mean value for k, N/m
8.73
K from the slope of the graph
8.72
% difference
.12%
0.6
Slide 18
Mass of the series of the two springs = 0.1805
Total suspended mass
m,kg
Period T,s
Frequency f, Square
Hz of period
T^2, s^2
K from
equation
(12), N/m
.3
1.456
.687
2.12
6,92
.35
1.552
.644
2.41
6.91
.4
1.641
.609
2.7
6.9
.42
1.676
.597
2.81
6.91
.45
1.725
.580
3
6.87
Slide 19
m versus T^2
T^2,
s^2
m, kg
0.5
y = 0.1714x - 0.0631
0.4
0.3
2.41
0.35
2.7
0.4
2.81
0.42
3
0.45
m, kg
2.12
0.3
0.2
0.1
0
0
1
2
3
T^2, s^2
Mean value for k, N/m
6.902
K from the slope of the graph
6.8
K from (17)
6.8
% Difference
1.5%
4
Slide 20
Conclusion:
After performing this experiment we can
conclude that k for the string is always
constant disregard to the mass of the
hanging object.
In addition, from the three graphs (m
versus T^2) for the strings and the series
we can observe the increasing function,
which means that the mass m is directly
proportional to the period of oscillation.
Slide 21
Understanding Problems:
A mass of 0.2 kg is attached to a spring with a force
constant k equal to 30N/m. if the mass executes
simple harmonic motion, what will be its frequency?
m = .2kg
k = 30N/m
= /2
= (k/m)
= (30/.2) = 12.25
= 12.25/2 = 1.95 Hz
Two springs with force constants of spring K1 and K2,
are connected in parallel. What is the spring constant
of the combination?
F = F1 + F2,
F = Kx
Kx = K1x1 + K2x2
x = x1 = x2,
K = K 1 + K2
Simple Harmonic
Motion:
Abdalla, Aymen
Spring 2002
SC.441.L.H, Sec (8773) Instructor: Dr. Roman
Kezerashvili
May – 17 - 2002
Slide 2
Introduction:
Periodic is any motion that
repeats it self in any equal
interval of time. A vibrating
spring and a simple
pendulum exhibit periodic
motion. The simple
harmonic motion is a
special type of periodic
motion.
Slide 3
Objectives
To study the simple harmonic
motion by investigating the
period of oscillation of a
spring, and to determine the
constant of the spring for one
spring and two springs in
series.
Slide 4
Equipments:
1.
2.
3.
4.
5.
Two springs.
Triple-beam balance.
Photo gate accessory.
Science workshop Interface box.
Set of masses.
Slide 5
Theory:
Consider a mass m, attached
to a spring. When the spring
in a stretched position, a
force F acts on the mass, and
x is the distance the mass
moves from its equilibrium.
This force tends to restore
the mass to its original
position and it is called the
restoring force. Also it is
opposite to the displacement
x and from hooks law.
Slide 6
F = -kx
(1)
K is the force constant of the spring.
From Newton’s second law :
a = F/m
Combine (1) and (2)
a = -kx/m or a= d^2x/dt^2 = -Kx/m (3)
(2)
Slide 7
But a varies with x, that:
^2 = k/m
or
= (k/m)
(4)
From equation (3)
A = -^2 x or d^2x/dt^2 = -^2x
(5)
And from the differential equation :
x = Acos(t + )
(6)
Slide 8
And from the differential equation
:
x = Acos(t + )
(6)
A is the maximum displacement
and it is called amplitude, = /2
is the frequency of vibration, is
the angular frequency. And (t +
) is called the phase of the simple
harmonic motion, is called the
phase constant.
Slide 9
For x from A to –A and back to A is
called a cycle ant T is the time for one
complete oscillation (cycle). That:
cos(t + + 2) = cos(t + )
So that:
(t + T) + = t + + 2 or t = 2(8)
So:
T = 2/
(7)
(9)
Slide 10
Sub from (4) to (9):
T= 2 m/k
(10)
This is true when the mass of the spring
ms is much less than the mass m, which is
suspended from the spring. And for a
spring of finite mass ms is:
T = 2 (m + ms/3)/k
(11)
From (11) determine k as:
k = 4^2(m + ms/3)/T^2
(12)
Slide 11
If we have two springs in series with the
force constant k1, k2 and x is the sum of
the displacement of each spring, that is:
x = x1 + x2
From Hook’s law:
x = F1/k1,
(13)
x2 = F2/k2, and x =F/k
Sub these values in (14):
F/k = F1/k1 + F2/k2
(15)
(14)
Slide 12
Because the springs in series
F = F1 = F2, that:
1/k = 1/k1 + 1/k2
or:
So that T is:
T = 2m(1/k1 + 1/k2)
(16)
k = k1k2/(k1+k2) (17)
(18)
Slide 13
Procedure:
In this experiment we measure the period of
oscillation, T for a spring with mass ms and mass
m for the object. Then use (12) to determine k,
and repeat the same for a series of two springs.
Use the science workshop to measure Frequency
and Number of cycles.
1. Measure the mass of the spring.
2. Suspend the mass of the object larger than the
mass of the spring.
3. Start record T period and frequency.
4. Increase the mass m and repeat step 4 for total
5-7 trails.
5. Connect two springs in series and suspend
mass m larger than the mass of the two springs,
and repeat steps 5 & 5.
Slide 14
Mass of the first single spring = 0.1805
Total suspended mass
m,kg
Period T,s
Frequency f, Square
Hz of period
T^2, s^2
K from
equation
(12), N/m
.2
1.075
.931
1.16
8.86
.25
1.19
.841
1.42
8.64
.3
1.28
.781
1.64
8.67
.35
1.363
.734
1.86
8.7
.4
1.44
.694
2.07
8.8
Slide 15
m versus T^2
m,
k
g
1.16
0.2
1.42
0.25
1.64
0.3
1.86
0.35
0.5
y = 0.2209x - 0.06
0.4
m, Kg
T^2,
s^2
0.3
0.2
0.1
0
0
2.07
0.5
0.4
1
1.5
2
T^2, s^2
Mean value for k, N/m
30.36
K from the slope of the graph
30.32
% difference
.13%
2.5
Slide 16
Mass of the second single spring = .035kg
Total suspended
mass m,kg
Period
T,s
Frequency Square
f, Hz of period
T^2, s^2
K from
equation
(12), N/m
.2
.524
1.91
.28
29.85
.25
.583
1.715
.34
30.5
.3
.638
1.567
.4
30.79
.35
.687
1.456
.47
30.5
.4
.734
1.362
.54
30.16
Slide 17
m versus T^2
m,
kg
0.28
0.2
0.34
0.25
0.4
0.47
0.3
0.35
0.5
0.4
m, kg
T^2,
s
^
2
y = 0.768x - 0.0118
0.3
0.2
0.1
0
0
0.2
0.4
T^2, s^2
0.54
0.4
Mean value for k, N/m
8.73
K from the slope of the graph
8.72
% difference
.12%
0.6
Slide 18
Mass of the series of the two springs = 0.1805
Total suspended mass
m,kg
Period T,s
Frequency f, Square
Hz of period
T^2, s^2
K from
equation
(12), N/m
.3
1.456
.687
2.12
6,92
.35
1.552
.644
2.41
6.91
.4
1.641
.609
2.7
6.9
.42
1.676
.597
2.81
6.91
.45
1.725
.580
3
6.87
Slide 19
m versus T^2
T^2,
s^2
m, kg
0.5
y = 0.1714x - 0.0631
0.4
0.3
2.41
0.35
2.7
0.4
2.81
0.42
3
0.45
m, kg
2.12
0.3
0.2
0.1
0
0
1
2
3
T^2, s^2
Mean value for k, N/m
6.902
K from the slope of the graph
6.8
K from (17)
6.8
% Difference
1.5%
4
Slide 20
Conclusion:
After performing this experiment we can
conclude that k for the string is always
constant disregard to the mass of the
hanging object.
In addition, from the three graphs (m
versus T^2) for the strings and the series
we can observe the increasing function,
which means that the mass m is directly
proportional to the period of oscillation.
Slide 21
Understanding Problems:
A mass of 0.2 kg is attached to a spring with a force
constant k equal to 30N/m. if the mass executes
simple harmonic motion, what will be its frequency?
m = .2kg
k = 30N/m
= /2
= (k/m)
= (30/.2) = 12.25
= 12.25/2 = 1.95 Hz
Two springs with force constants of spring K1 and K2,
are connected in parallel. What is the spring constant
of the combination?
F = F1 + F2,
F = Kx
Kx = K1x1 + K2x2
x = x1 = x2,
K = K 1 + K2