Measurement of Elastic Constant of Spiral Spring, and Earth's Gravitational Intensity
of
Elastic Constant of Spiral Spring, and Earth's Gravitational Intensity
. Elastic Constant Of Spring
idea is to investigate the
relationship for the extension of spiral spring x and for the oscillation
period of this spring T of suspended different mass m.a mass is suspended from
the end of a spring, Hooke's Law states that that the extension of the spring
is proportional to the mass provided the elastic limit of the spring is not
exceeded., the tension force, F (N), in the spring is proportional to
the extension x (m) produced. That is:
= kx,
k (N/m) is the spring (elastic)
constant.
a mass m is placed on the spring,
acting on the mass gravitational field of the Earth. Then the spring is
stretched until the force of gravity mg is balanced by the elastic force of the
spring F (Fig.1). And given by
= mg = kx.(1)
the tension of the spring is then
mg., by measuring the mass of the load m and the corresponding extension x
(OB, Fig.1) of the spring, it is possible to find the ratio the acceleration
due to gravity at this point on the Earth's surface by the spring constant this
spiral spring (k/g).equation (1):
,(2)
:= m + m0 - total mass of
load on spring, which stretches a spring (kg), m - the total mass of load (kg),
m0 - the mass of holder (kg);- extension of spring, mean average
reading length spring minus reading initial length spring x=l - l0).k/g
- the gradient a graph of extension x against the mass M (kg/m)., we can plot a
graph of extension x against the mass M whose weight extended the spring to
found the ratio the acceleration due to gravity at this point on the Earth's
surface by the spring constant this spiral spring (k/g).this we must draw the
best straight line through the origin. From the graph calculate the gradient =
(k/g).Variable: mass of load on spring (m).Variable: length spring after
stretches (l).Variables: initial length spring l0; mass of holder m0
.
. Earth Gravitational Intensity
(g)
the mass is pulled down a little and
then released, it oscillates up and down above and below O (position of equilibrium)
(Fig1, b).true the motion is simple harmonic about O, and the period
oscillation of spring is given by:
. (3)
T - period of oscillation of the
spring (s);- total mass of load on spring (kg);- where m is a constant
depending on the mass of the spring itself;- elastic constant (N/m).(3), it
follows that:
) = .(4)
of oscillation of the spring T we
found in this way:
=t/N,
:- time taken N oscillation (s);-
number of oscillation.a graph of T2 against M should
be a straight line and the y-intercept is m.of this graph gives us:
= .
we can found elastic constant k for
this spiral spring.the value of acceleration due to gravity at this point on
the Earth's surface can be found by substituting the magnitudes of gradient at
part 1 and elastic constant k in part 2:
g = k/ gradient.
data:
Zero reading of the spring (initial
length of the spring) l0 = 0,02 m.holders B to the spring m0
= 0,049 kg.
Error:
To measure the initial length and
length of the spring after extension I used a ruler, the absolute uncertainty
of which was equal to ± 0,5 mm, because the limit uncertainty metre rule is
half the limit of reading (1 mm).of holder was measured on a digital balance
with a precision of ± 0.1 gmass of load was measured on a digital balance with
a precision of 0.1 g. In all cases the measured mass was less than 1 g off. A
typical measure is m1 = 199.3 g. The masses are thus assumed
to be accurate to ±1 g or Δm = ± 0.001kg.measure
the time of 10 complete oscillations I used stopwatch, the absolute uncertainty
of which was equal to ± 0.1 s.as to obtain more accurate data, I measured time
of 10 complete oscillations by load decreasing for same mass of load three
times.
1. The measurement results of load
mass and reading on ruler
Raw Data Measure
|
Mass of load m/ (kg) Δm±
0,001 kg
|
Reading on ruler by load increasing l / (m) Δl
= ±0,005 m
|
1
|
0,050
|
0,042
|
2
|
0,100
|
0,056
|
3
|
0,150
|
0,071
|
4
|
0,200
|
0,085
|
5
|
0,250
|
0,099
|
6
|
0,300
|
0,114
|
7
|
0,350
|
0,128
|
8
|
0,400
|
0,142
|
9
|
0,450
|
0,156
|
10
|
0,500
|
0,171
|
Table 2. The measurement results of
total mass of load and time of 10 complete oscillations
Raw Data Measure
|
Mass of load m/ (kg) Δm±
0,001 kg
|
Trial# 1
|
Trial# 2
|
Trial# 3
|
|
|
Time of 10 complete oscillations by load
decreasing t (s) Δt = ±0,1
s
|
Time of 10 complete oscillations by load
decreasing t (s) Δt =±0,1
s
|
1
|
0,500
|
8,3
|
8,3
|
8,2
|
2
|
0,450
|
8,0
|
8,1
|
8,1
|
3
|
0,400
|
7,5
|
7,6
|
7,7
|
4
|
0,350
|
7,3
|
6,1
|
7,3
|
5
|
0,300
|
6,8
|
6,6
|
6,6
|
6
|
0,250
|
6,4
|
6,1
|
6,3
|
7
|
0,200
|
5,6
|
5,7
|
5,8
|
8
|
0,150
|
5,0
|
5,1
|
5,1
|
9
|
0,100
|
4,6
|
4,8
|
4,9
|
10
|
0,050
|
3,9
|
4,1
|
3,8
|
Calculations
spiral spring
gravitational intensity
Table 3. Measurement error and
calculations
Total mass of load M / (kg) ΔM±
0,002 kg ± 0,6 %.
|
Extension x / (m) Δl
= ±0,002 m ±1,0 %
|
Average time of 10 complete oscillations t (s)
± 1,0 %.
|
Period of oscillation T (s) ± 2%.
|
Oscillation period in the square T2
(s2) ± 4%.
|
0,549
|
0,022
|
8,3
|
0,83
|
0,68
|
0,499
|
0,036
|
8,1
|
0,81
|
0,65
|
0,449
|
0,051
|
0,76
|
0,58
|
0,399
|
0,065
|
6,9
|
0,69
|
0,48
|
0,349
|
0,079
|
6,7
|
0,67
|
0,44
|
0,299
|
0,094
|
6,3
|
0,63
|
0,39
|
0,249
|
0,108
|
5,7
|
0,57
|
0,32
|
0,199
|
0,122
|
5,1
|
0,51
|
0,26
|
0,149
|
0,136
|
4,8
|
0,48
|
0,23
|
0,099
|
0,151
|
3,9
|
0,39
|
0,15
|
for total mass of load M:: (0,549 +
0,099)/2 = 0,324 kg.uncertainty:0,002/0,325 = ± 0,006.uncertainty:0,006*100% =±
0,6 %.for extension x::(0,151 + 0,022)/2 = 0,087 m.uncertainty:0,001/= 0,087 =
± 0,011.uncertainty:0,011 * 100 % = ± 1,0 %for period of oscillation T:: (0,83
+ 0,39)/2 = 0,61 s.uncertainty:0,01/0,61 = ± 0,02.uncertainty:0,02*100% =± 2,0
%.for oscillation period in the square T2:2,0 % + 2,0 % = 4,0 %graph
of Oscillation period in the square against of Total mass of load:gradient:
(0,099 + 0,6 % = 0,100 kg);(0,15 - 4,0 % = 0,14 s2).
(0,549 - 0,6 % = 0,546 kg);(0,68 +
4,0 % = 0,71 s2).
gradient: (0,099 - 0,6 % = 0,098
kg);(0,15 + 4,0 % = 0,16 s2).
(0,549 + 0,6 % = 0,552 kg);(0,68 -
4,0 % = 0,66 s2).
graph of Total mass of load against
Extensiongradient: (0,022 + 1,0 % = 0,022 m);(0,099 - 0,6 % = 0,098 kg).
(0,151 -1,0 % = 0,150 m);(0,549 +
0,6 % = 0,552 kg).
gradient: (0,022 - 1,0 % = 0,022
m);(0,099 + 0,6 % = 0,100 kg).
(0,151 + 1,0 % = 0,152 m);(0,549 -
0,6 % = 0,546 kg).
Presenting processed data:
Part 2. To process the data, I plot
the oscillation period in the square T2 of total mass of load M
(Fig.3). The graph, I have shown a best fit line and its equation.graph I
plotted the error bars for oscillation period in the square T2 and
total mass of load.
.2
I have plotted the graph of maximum and
minimum gradient., we can calculate the value of the spring constant of spring
of gradient obtained from the graph:
= =
34,79 Nm-1.
of uncertainty in this value of the
spring constant of spring using max / min gradients:
kmin = =
32,30 Nm-1kmax = =
34,53 Nm-1 ≈ 34,90 Nm-1
of uncertainty in this value of the
spring constant of spring using max / min gradients:
min
= =
31,16 Nm-1kmax = =
35,92 Nm-1 .
the absolute uncertainty in this
value of the spring constant of spring:
Δk = ± =
=
±2,38 Nm-1
the experimental value and absolute
uncertainty spring constant:
exp
± Δk ≈
(34,79 ± 2,38) Nm-1
the fractional uncertainty in the
spring constant of spring: Fractional uncertainty = (2,38)/(34,79) = 0,068 ≈
0,07. Calculate the percentage uncertainty in this value of the resistivity of
constantan:
uncertainty = 0,07 ×
100% = 7,0 %.
, we can calculate the value of
acceleration due to gravity of gradient obtained from the graph:
= =
9,95 Nkg-1 .
of uncertainty in this value of the
acceleration due to gravity using max / min gradients:
min
= =
9,85 Nkg-1gmax = =
10,06 Nkg-1
the absolute uncertainty in this
value of the acceleration due to gravity:
Δg = ± =
=
± 0,11 Nkg-1
the experimental value and absolute
uncertainty acceleration due to gravity:
g exp ± Δg
≈ (9,95 ± 0,11) Nkg-1
the fractional uncertainty in the
acceleration due to gravity:
the percentage uncertainty in this
value of acceleration due to gravity:
uncertainty = 0,01 ×
100% = 1,0 %.
must say that the calculation does
not account for the gradients of the error for the spring constant of spring ,
total mass of load and the extension of spring.of percentage uncertainty in
this value of the acceleration due to gravity using uncertainty for the spring
constant of spring , total mass of load and the extension of spring:
uncertainty :7 % + 0,6 % + 1,0 % =
7,6 % 8
%.
the absolute uncertainty of the
acceleration due to gravity:
Δ g = 9,95 x 0,08 = 0,796 0,80
Nkg-1
the experimental value and absolute
uncertainty acceleration due to gravity:
exp
± Δg ≈
(9,95 ± 0,80) Nkg-1
I believe that the result is more
correct to assess of the acceleration due to gravity.
Concluding and Evaluation:
My graphs support the theory because
it is a straight line, but which not passes through the origin. This suggests a
systematic experimental error. I believe that the systematic error due to the
fact that the spring, first, already had initial stretching. This systematic
error is clearly visible on Fig.4.we look at the Fig.2, we can say that the
systematic error and with the further due to the fact that we did not measure
the period of oscillation of the spring holder only., according to equation (3)
y-intercept is m. So can be
determined from the graph (Fig.2) the mass of the spring itself:
,029 = ;=
=
0,026 ≈( 30 ± 1) g.
error observed in the graph (Fig.2),
since the best-fit line does not pass all the errors bars. Clearly, the source
of greatest error in the experiment is in the measurement of the period. So,
the error total mass of load is 0,6 %, extension is 1,0 %, whereas the error
square period of oscillation to 4,0 %.calculation results are given for in the
spring constant of spring is 7,0 % and percentage uncertainty in this
experimental value of the acceleration due to gravity is 8,0 %.different points
on Earth, objects fall with acceleration between 9.78 and 9.82 ms-2
with a conventional standard value of exactly 9.80665 ms-2.conclusion
the experimental range is from 9,15 Nkg-1 to 10.75 Nkg-1
and this range includes the accepted value.can assume that it was the influence
of oscillation in air resistance, since the form of load was flat. And may have
been loss of energy and oscillation were slightly, but damped. Besides,
possibly, human error in the measurement was an appropriate amount of time and
oscillation.is also a reliance on other data that is assumed rather than
measured such as the constant depending on the spring itself and hence giving
inaccurate readings.graphs show that significant anomalies and emissions during
the experiment are not received.
Improving
The slight but noticeable scatter of
data about the best straight-line could be improved by taking more readings.
For example, use a fractional mass of load. You can also add measurements in
reverse order, that is, to reduce the mass of the load., repeated readings
there is a possibility that the spring will have a residual extension.it is
possible to propose to use in this experiment, a spring with a large spring
constant in order to more reduce damping. And use load which will have a more
streamlined shape, in order to reduce the influence of air resistance.