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Bifilar Suspension - the technique will be applied to find the mass moment of inertia of a regular cross-section steel beam about its centre of gravity.

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Introduction

´╗┐Suspension bridges : AIM OF THE EXPERIMENT: To find the relation between the time take and the distance in a suspension bridge where the distance increases as (6cm, 8cm, 10cm, 12cm, 14cm, 16cm.) INTRODUCTION: The Bifilar Suspension is a technique that could be applied to objects of different shapes, but capable to be suspended by two parallel equal-length cables, in order to evaluate its mass moment of inertia I about any point within the body. In this experiment, the technique will be applied to find the mass moment of inertia of a regular cross-section steel beam about its centre of gravity. This is done by suspending two parallel cords of equal length through the object examined. The system oscillates because any rotation of the rod away from its equilibrium position moves the strings away from the vertical, which lifts the rod upwards and increases its potential energy. VARIABLES: Independent variables: The distance from the middle point of both parallely arranged scles i.e. 50cm and 25cm. The distance starts from 6cm, 8cm, 10cm, 12cm, 14cm and 16cm on both th sides. The initial distance was chosen to be 6cm because the measures below that did not favour the setup as the oscillations were not possible. Dependent variables: The time taken for 10 oscillations by the scale which was measured using a digital stop watch. ...read more.

Middle

* The values below 12 cm did not favour the experiment because the ruler’s oscillations were not possible. OBSERVATION TABLE: S.no Distance from the mid-point in cm (± 0.05cm) Time taken for 10 oscillations in s (± 0.5 s) Tmean Trial 1 Trial 2 Trial 3 1 13 11 11 11 11 2 16 10 10 10 10 3 19 09 09 09 09 4 22 08 08 08 08 5 25 07 07 07 07 6 28 06 06 06 06 PROCESSED DATA ANALYSIS: S.no Distance from the mid-point in cm (± 0.05cm) Time for 10 oscillations in s (± 0.5s) Time/ 10 (1/Time2) k = 2 Δ k 1 13 11 1.1 0.83 21.73 7.36 2 16 10 1.0 1.00 25.12 3.97 3 19 09 0.9 1.23 28.07 1.01 4 22 08 0.8 1.56 30.77 1.68 5 25 07 0.7 2.04 33.28 4.20 6 28 06 0.6 2.78 35.54 6.46 avg = 29.09 avg = 4.12 RANDOM ERROR: = (Δ k avg/ k avg) x 100 % = (4.12/ 29.09) x 100 % = 14.13 % GRAPH: 1 Graph analysis: * The graph plotted above is between effective distance (D) and time period (T) * The line is not passing through origin (0,0) indicating the presence of systematic error in the investigation. ...read more.

Conclusion

/ 0.09429X 100 = 9.566 % CONCLUSION: * From the experiment conducted it is found that it is true and that as the distance increases the time period decreases. * The straight line of the linear graph shows that there is a linear relationship between the mass of the distance and time period. * The y-intercept shows the error that exists in the experiment. The y-intercept is -0.051010 cm which is very low. * From the graph it is found that k = 2 * The best fit line shows the accuracy of the experiment as the line coincides with most of the points and passes through the all the error bars. The true value of spring constant is * The accuracy depends upon the random error and as the random error is less that is 14.13% the experiment is more accuarate. * Maxima-Minima graph shows the accuracy of the readings taken and the slope of the lines. * The analysis of uncertainties and errors has been done and it is found that the systematic error is less. * The random error is 12.13% and the systematic error is 9.566%. Evaluation: * The distance is taken till 32cm till the oscillations were possible. * The human parallax error also accounts the reading. * The string was not perfectly inextensible. * The zero error may also effect the experiment. ...read more.

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