Measurement of Elastic Constant of Spiral Spring, and Earth's Gravitational Intensity

Investigation the relationship for the extension of spiral spring and for the oscillation period of this spring of suspended different mass. Earth Gravitational Intensity. The results of load mass and reading on ruler. Measurement error and calculations.

Рубрика Производство и технологии
Вид реферат
Язык английский
Дата добавления 17.10.2014
Размер файла 209,9 K

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Measurement of Elastic Constant of Spiral Spring, and Earth's Gravitational Intensity

1. Elastic Constant Of Spring

The 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.

When 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.

Generally, the tension force, F (N), in the spring is proportional to the extension x (m) produced. That is:

F = kx,

where k (N/m) is the spring (elastic) constant.

When 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

F = mg = kx.(1)

Since the tension of the spring is then mg.

Thus, 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).

From equation (1):

,(2)

where:

M = 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);

x - extension of spring, mean average reading length spring minus reading initial length spring x=l - l0).

Therefore k/g - the gradient a graph of extension x against the mass M (kg/m).

So, 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).

For this we must draw the best straight line through the origin. From the graph calculate the gradient = (k/g).

Independent Variable: mass of load on spring (m).

Dependent Variable: length spring after stretches (l).

Controlled Variables: initial length spring l0; mass of holder m0 .

2. Earth Gravitational Intensity (g)

If the mass is pulled down a little and then released, it oscillates up and down above and below O (position of equilibrium) (Fig1, b).

A true the motion is simple harmonic about O, and the period oscillation of spring is given by:

. (3)

Where T - period of oscillation of the spring (s);

M - total mass of load on spring (kg);

m - where m is a constant depending on the mass of the spring itself;

k - elastic constant (N/m).

From (3), it follows that:

) = .(4)

Period of oscillation of the spring T we found in this way:

T=t/N,

where:

t - time taken N oscillation (s);

N - number of oscillation.

Consequently a graph of T2 against M should be a straight line and the y-intercept is m.

Gradient of this graph gives us:

grad = .

Therefore we can found elastic constant k for this spiral spring.

Finally 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.

Raw data:

Zero reading of the spring (initial length of the spring) l0 = 0,02 m.

Mass 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).

Mass of holder was measured on a digital balance with a precision of ± 0.1 g

Each mass 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 Дm = ± 0.001kg.

To measure the time of 10 complete oscillations I used stopwatch, the absolute uncertainty of which was equal to ± 0.1 s.

And as to obtain more accurate data, I measured time of 10 complete oscillations by load decreasing for same mass of load three times.

Table 1. The measurement results of load mass and reading on ruler

Raw Data

Measure

Mass of load

m/ (kg)

Дm?????± 0,001 kg

Reading on ruler by load increasing

l / (m)

Дl = ±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)

Дm????± 0,001 kg

Trial# 1

Trial# 2

Trial# 3

Time of 10 complete oscillations by load decreasing

t (s)

Дt = ±0,1 s

Time of 10 complete oscillations by load decreasing

t (s)

Дt =±0,1 s

Time of 10 complete oscillations by load decreasing

t (s)

Дt =±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)

ДM?????± 0,002 kg

± 0,6 %.

Extension

x / (m)

Дl = ±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

7,6

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

Error for total mass of load M:

Average: (0,549 + 0,099)/2 = 0,324 kg.

Fractional uncertainty:0,002/0,325 = ± 0,006.

Percentage uncertainty:0,006*100% =± 0,6 %.

Error for extension x:

Average:(0,151 + 0,022)/2 = 0,087 m.

Fractional uncertainty:0,001/= 0,087 = ± 0,011.

Percentage uncertainty:0,011 * 100 % = ± 1,0 %

Error for period of oscillation T:

Average: (0,83 + 0,39)/2 = 0,61 s.

Fractional uncertainty:0,01/0,61 = ± 0,02.

Percentage uncertainty:0,02*100% =± 2,0 %.

Error for oscillation period in the square T2:2,0 % + 2,0 % = 4,0 %

For graph of Oscillation period in the square against of Total mass of load:

Max 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).

Min 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).

For graph of Total mass of load against Extension

Max gradient: (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).

Min 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.

In graph I plotted the error bars for oscillation period in the square T2 and total mass of load.

Fig.2

Also I have plotted the graph of maximum and minimum gradient.

So, we can calculate the value of the spring constant of spring of gradient obtained from the graph:

k = = 34,79 Nm-1.

Calculation 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

Calculation of uncertainty in this value of the spring constant of spring using max / min gradients:

kmin = = 31,16 Nm-1kmax = = 35,92 Nm-1 .

Calculate the absolute uncertainty in this value of the spring constant of spring:

Дk = ± = = ±2,38 Nm-1

Therefore the experimental value and absolute uncertainty spring constant:

kexp ± Дk ? (34,79 ± 2,38) Nm-1

Calculate 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:

percentage uncertainty = 0,07 Ч 100% = 7,0 %.

So, we can calculate the value of acceleration due to gravity of gradient obtained from the graph:

g = = 9,95 Nkg-1 .

Calculation of uncertainty in this value of the acceleration due to gravity using max / min gradients:

g min = = 9,85 Nkg-1gmax = = 10,06 Nkg-1

Calculate the absolute uncertainty in this value of the acceleration due to gravity:

Дg = ± = = ± 0,11 Nkg-1

Therefore the experimental value and absolute uncertainty acceleration due to gravity:

g exp ± Дg ? (9,95 ± 0,11) Nkg-1

Calculate the fractional uncertainty in the acceleration due to gravity:

Fractional uncertainty = (0,11)/(9,95) = 0,01

Calculate the percentage uncertainty in this value of acceleration due to gravity:

Percentage uncertainty = 0,01 Ч 100% = 1,0 %.

I 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.

Calculation 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:

Percentage uncertainty :7 % + 0,6 % + 1,0 % = 7,6 % 8 %.

So the absolute uncertainty of the acceleration due to gravity:

Д g = 9,95 x 0,08 = 0,796 0,80 Nkg-1

Therefore the experimental value and absolute uncertainty acceleration due to gravity:

g exp ± Дg ? (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.

If 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.

Since, according to equation (3) y-intercept is m. So can be determined from the graph (Fig.2) the mass of the spring itself:

0,029 = ;

m = = 0,026 ?( 30 ± 1) g.

Random 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 %.

The 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 %.

At 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.

In conclusion the experimental range is from 9,15 Nkg-1 to 10.75 Nkg-1 and this range includes the accepted value.

We 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.

There 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.

The 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.

Nevertheless, repeated readings there is a possibility that the spring will have a residual extension.

Besides 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.

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