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Bending Moment Experiment

By davidboadu Dec 10, 2010 1762 Words
TOPIC:
Spring Mass Oscillator
OBJECTIVE:
To determine the spring constant (K), using mass system.

APPARATUS:

STEEL RULE
SPRING

STOP WATCH

TAPE MEASURE
SLOTTED MASS

THEORY:
In classical mechanics, a harmonic oscillator is a system which , when displaced from its equilibrium position, experience a restoring force, F, proportional to the displacement, X, according to Hooke’s Law; F = – KX = mα …………………………………. Where,

F = the restoring force used to displace the mass. K= the spring constant.
X = the displacement of the mass with respect to the equilibrium. The most commonly encountered form of Hooke’s Law is probably the spring equation, which relates the force exerted by a spring to the distance it is stretched by a spring constant, K, measured in force per length. The negative sign indicate that the force exerted by the spring is in direct opposition to the direction of displacement. The force is called a ‘restoring force’, as it tends to restore the system to equilibrium. Considering the diagram below;

(a) (b) (c)

K K K

∂ + x
+x m x
m
(a) The natural length of the spring.
(b) Equilibrium position of the spring mass system.
(c) The system displaced a small distance, ∂, from the equilibrium position.
If the mass is given a small vertical displacement, x, from its equilibrium position, it will oscillate about the equilibrium position.
In the equilibrium condition;
Weight downwards = Spring restoring force upwards
i.e. mg = K∂ …………………………. (1)
When the mass is pulled down beyond its equilibrium position by a distance, x , there is a residue of force available to accelerate the mass.
From Newton’s Second Law of motion;
mα = mg – K (∂ + x )
mα = mg – k∂ – Kx ……………………… (2 )
But from equation (1)
mg = K∂
mα = mg – mg – Kx
mα = – Kx
α = –
Where, ω2 = –
As ω2 is constant the acceleration is proportional to the displacement, x and the negative sign indicates that it is directed in the opposite direction to x .
ω2 =
ω =
Also Period of oscillation, T, = 2π ω
Since, ω =
T = 2π …………………….(3)
Making K the subject of the equation,
T2 = ……………………..(b)
K = ……………………(4)

METHOD OR PROCEDURE FOR THE FIRST EXPERIMENT (STATIC)
* Hang the spring from the support end and place a weight hanger. * Measure the height from the bottom of the weight hanger to the top of the spring. * Suspend 1N weight on the hanger and measure the height. * Continue with the increment of 1N until 5N weight is reached. * Plot a graph of F Vs ∆x, where F is the weight hanging from the spring and ∆x is the displace caused by the weight. * Determine the spring constant which is the slop of the best – fit line of the graph.

RECORDING AND TABULATING OF RESULTS
S/N| HEIGHT OF WEIGHT HANGER AND SPRING | FORCE | HEIGHT OF WEIGHT HANGER, SPRING AND LOAD | DEFLECTION | 1| 0.433| 1| 45.1 x 10-2| 0.451 - 0.433|
2| 0.433| 2| 46.4 x 10-2 | 0.464 - 0.433 |
3| 0.433| 3| 47.9 x 10-2| 0.479 - 0.433|
4| 0.433| 4| 48.9 x 10-2| 0.489 - 0.433|
5| 0.433| 5| 50.2 x 10-2| 0.502 - 0.433|

SAMPLE CALCULATIONS

(1) (2)
= 0.451 – 0.433 = 0.464 – 0.433
= 0.018m = 0.31m

(3) (4) = 0.479 – 0.433 = 0.489 – 0.433

= 0.046m = 0.056m

TABLE OF DERIVED RESULTS FOR THE FIRST EXPERIMENT (STATIC)
S/N| HEIGHT OF WEIGHT HANGER AND SPRING | FORCE | HEIGHT OF WEIGHT HANGER, SPRING AND LOAD | DEFLECTION | 1| 0.433| 1| 0.451| 0.018|
2| 0.433| 2| 0.464| 0.031|
3| 0.433| 3| 0.478| 0.046|
4| 0.433| 4| 0.489| 0.056|
5| 0.433| 5| 0.502| 0.069|
Graph of F against x

Graph sheet 1

Graph of F against ∆x
From equation;

But from the graph;
Gradient

METHOD OR PROCEDURE FOR THE SECOND EXPERIMENT (MOTION)
* Firstly measure the mass of the spring.
* Place 15N weight on the weight hanger and set the spring to oscillate by pulling the weight hanger down and releasing it. * Record the time (period ‘T’ for 20 oscillation) taken for the apparatus to complete the oscillation. * Repeat cycle with increment of 5N weight until 40N weight is reached. * Plot the graph of T2 Vs m (this mass consisting of the hanging mass plus of the mass of the spring) * Determine the slop of the best – fit line which is the spring constant, K.

RECORDING AND TABULATING OF THE RESULTS
S/N| WEIGHT,(N)| MASS OF SPRING (ms)| TIME FOR 20 OSC.| TOTAL PERIODIC TIME| AVERAGE TIME| AVERAGE TIME/20| PERIOD,T2| Mass(m)mh +1/3ms| | | | t1| t2| t3| | | | | |
1| 15| 0.118| 17.50| 17.47| 17.40| 52.37| 17.46| 0.873| 0.762| 1.568| 2| 20| 0.118| 19.91| 19.93| 19.97| 59.81| 19.94| 0.997| 0.994| 2.078| 3| 25| 0.118| 22.22| 22.15| 22.19| 66.56| 22.186| 1.109| 1.229| 2.587| 4| 30| 0.118| 24.07| 24.12| 24.06| 72.25| 24.08| 1.204| 1.450| 3.097| 5| 35| 0.118| 25.65| 25.45| 25.52| 76.62| 25.54| 1.277| 1.631| 3.607| 6| 40| 0.118| 27.63| 27.50| 27.88| 83.01| 27.67| 1.384| 1.915| 4.116|

SAMPLE CALCULATIONS
Mass of spring (ms) = 0.118kg
But ω = mg
m =
Where, g = 9.81m/s2
(1) mh1 =
= 1.529kg
Mass (m) =
M1 = = 1.568kg
(2) mh2 = = 2.039kg (3) mh3 = = 2.548kg
Mass (m) = Mass (m) =
m2 = m3 =
= 2.079kg = 2.587kg

(4) mh4 = = 3.058kg (5) mh5 = = 3.568kg
Mass (m) = Mass (m) =
m3 = m5 =
= 3.097kg = 3.607kg

TABLE OF DERIVED RESULTS FOR THE SECOND EXPERIMENT (MOTION) S/N| WEIGHT,(N)| MASS OF SPRING (ms)| PERIODIC TIME (T)| | T2| Mass(m)mh +1/3ms| 1| 15| 0.118| 17.46| 0.873| 0.762| 1.568|

2| 20| 0.118| 19.94| 0.997| 0.994| 2.078|
3| 25| 0.118| 22.25| 1.113| 1.239| 2.587|
4| 30| 0.118| 24.08| 1.204| 1.450| 3.098|
5| 35| 0.118| 26.11| 1.306| 1.706| 3.607|
6| 40| 0.118| 27.67| 1.384| 1.915| 4.116|

Graph sheet 2

Graph of T2 against m

From equation (b);
T2

Where, = Gradient (Mx)

From the graph;
Gradient (Mx)

PRECAUTIONARY MEASURES TAKEN
* When measuring the length of the spring and the weight hanger, the apparatus was ensured that it was not oscillating. This was done to ascertain the right length of reading. * It was ensured that the displacement of the spring and its timing for 20 oscillations was achieved perfectly and accordingly, starting from zero to twenty. * When setting the spring to oscillate, it was ensured that the length of displacement was not more than the length of the mass on the weight hanger. * The spring was ensured that it was well secured before it was set into motion.

RECOMMENDATION
* The laboratory should be regularized to prevent unauthorized persons from entering. * The laboratory should be well stuffed with modern equipment and apparatus. * The equipment and apparatus in the laboratory should be properly arranged to meet modern standards. * The laboratory should not be used for lecture hall since it creates inconveniences * Laboratory supervisors should discuss lab reports or results of an experiment with students so that they do not repeat errors the next time they are asked to experiment again. * The number of the apparatus should be increased to facilitate earlier submission of the report.

CONCLUSION
A spring constant is a measure of how stiff a spring is. A spring that is very hard to stretch out has a large spring constant whilst a spring which is easy to stretch has a small spring constant. As the term spring constant implies, the spring constant is always the same for a given spring, assuming you do not put so much force on it that you break it.

The experiment, in determining the spring constant was carried out satisfactorily. It could be concluded that;
the SI standard values for the spring constant, K from mechanics of machines book when compared to the value obtained from the experiment, could be stated that the objective of the experiment was achieved to some extent.

However, for some length of values which were little below or above which was obtained from the experiment could be attribute to minor errors that may have been encountered during the experiment.

It is believed that, the precautionary measures if abide by them any time such an experiment is given, the standard values would perhaps be obtained or even nearly to some degree of accuracy.

QUESTION
How is the value of the spring constant related to the stiffness of the spring?
BEST ANSWER
Stiffness is a measure of resistance to deformation.

The spring constant is related to the slope of the stress/strain curve in and around the region in which the spring exhibits a linear elastic response.

The spring constant is defined as k where F = kx. So the bigger k, the more force (compression/tension) required to achieve a unit deformation (x).

A stiffer spring has a shallower slope (greater spring constant) because more stress (compression/tension) is required per unit deformation (strain.)

QUESTION
A mass m1 is suspended from a fixed point by means of a spring of stiffness S1. Attached to m1 by means of a second spring of stiffness S2 is another mass m2 when the system is set in free vibration, if the inertia of the springs be neglected. Show that

where ω is the phase velocity. If m1 and m2 are each 225kg and S1 and S2 are 240 and 120 KN/m respectively. Find the frequencies of oscillation and the ratio of the amplitude of m1 and m2.

S1

m1
χ1
S2

m2
χ2

SOLUTION
Let the instantaneous displacement of m1 and m2 be χ1 and χ2 respectively, then the equation of the motion of m1 and m2 =
…………………… (1)
……………………..(2)
Let χ1 = α1 cosωt …………….(3)
χ2 = α2 cosωt ………….. (4)
From equation (3)

………………………(5)
Substituting equations (3) and (5) into equation (1)
gives; ……………… (6)
Dividing through the equation by

……………………………… (7)

From equation (4)

………………………… (8)

Substituting equations (4) and (8) into equation (2)
gives; …………………….. (9)

Divide through the equation by

………………………. (10)
Equating equations (7) and (10)

……………………… (11)
From equation (11)

Divide through the equation by 50625,
gives;
Let

0r

From equation (7), when

From equation (10), when

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