# 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)

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.

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