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Helical Spring Experiment (Class 11) Readings

Helical Spring Experiment with Readings (Class 11)

Experiment - 1


Aim: To find the spring constant k of a helical spring by measuring time period of vertical oscillations of a known load.


Apparatus: A helical spring, a rigid support, metre scale, weight hanger, a set of equal weights (20 g or 50 gm slotted weights), a fine pointer, stop watch, etc.


Procedure: The helical spring is balanced vertically from a rigid support. The pointer is attached horizontally at the free end of the spring. A 20 g (or 50 g) weight hanger is suspended from the hook at the free end of the spring. The weight m of the hanger should be enough to keep the spring vertical and stretched. Now the load M at the end of the spring is lm. The metre scale is set aside vertically such that its tip of the pointer comes over the divisions on the scale; also confirm that it is not touched the scale. The weight hanger is pulled a bit downwards and left free. The load moves up and down executing vertical oscillations. Time t for, say; 20 oscillations is determined. From this, the period of oscillations, T = t/20, of the spring is calculated. Then (M/T2) is calculated.


A slotted weight of mass m is placed on the weight hanger. Now the load M at the end of the spring is 2m. As before, (M/T2) is calculated. The experiment is repeated for the load M equals to 3m, 4m, etc. The average value of (M/T2) is determined. The spring constant k is calculated from the equation,


k = 4π2(M/T2)


The spring constant can also be calculated graphically as mentioned below. A graph is plotted with load M (1m, 2m, 3m, etc.) in kg along the X-axis and T2 along the Y-axis. The graph is a straight line. The reciprocal of the slope of the graph i.e., (M/T2) is determined. The spring constant of the spiral spring is calculated from the equation, k = 4π2(M/T2)


Observations and Readings


To find k by Vertical Oscillations,


Slotted weight, m = 20 gwt


Load M x 10-3 kg

Time for 20 Oscillations

Period, T(s), t/20

T2

M/T2

1

2

Mean t(s)

30 x 10-3

13

12

12.5

0.62

0.38

0.078

40 x 10-3

14

14

14

0.7

0.49

0.081

50 x 10-3

15

15

15

0.75

0.56

0.089

60 x 10-3

15

15

15

0.75

0.56

0.107

70 x 10-3

16

16

16

0.8

0.64

0.109

 

Mean (M/T2) = 0.092 kg/s2

K = 4π2(M/T2) = 3.86 Nm-1

From the graph,

AB = 27.5 x 10-2; BC = 0.30

Therefore, M/T2 = AB/BC = 0.092

K = 4π2(M/T2) = 3.62 Nm-1

Mean K = 3.62 Nm-1


Result


Spring constant, k, by vertical oscillations = 3.62 Nm-1


Experiment 2:


Aim: To find the spring constant k of a helical spring from load-extension graph.


Observations and Readings:


Reading of the pointer for the dead load W0 m

(i) Loading = 24.7 cm

(ii) Unloading = 24.7 cm

Mean r0 = 24.7 cm


Load, M x 10-3 kg

Reading of the pointer

Extension e (cm)

Extension e (m)

M/e = K (Kg/m)

Loading

Unloading

Mean

10 x 10-3

23.7

23.7

23.7

1

1 x 10-2

10 x 10-1

20 x 10-3

22.6

22.6

22.6

2.1

2.1 x 10-2

9.52 x 10-1

30 x 10-3

21.6

21.6

21.6

3.1

3.1 x 10-2

9.57 x 10-1

40 x 10-3

20.8

20.8

20.8

3.9

3.9 x 10-2

10.25 x 10-1

50 x 10-3

19.7

19.7

19.7

5

5 x 10-2

10 x 10-1

60 x 10-3

18.6

18.6

18.6

6.1

6.1 x 10-2

9.83 x 10-1

70 x 10-3

17.5

17.5

17.5

7.2

7.2 x 10-2

9.72 x 10-1

 

a) Spring Constant K


by Calculation,

K = (M/e) = 0.98 kgwt/m = 9.604 N/m-1


b) From the Graph,


AB = 40 x 10-3 kgwt

BC = 0.040 m

K = AB/BC = 1 Kgwt m-1 = 1 x 9.8 = 9.8 Nm-1

 

Result


Spring constant (a) by calculation = 9.604 N/m-1

(b) k, from load-extension graph = 9.8 Nm-1

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