I feel that if I increase the mass, the period will increase too because the mass is proportionate to the time period.Variables> Independent – Mass suspended from spring.> Dependent- The time period of the string.> Constant- The apparatus, Number of oscillations, spring constant (elasticity).Apparatus> 1 spring> 100 gram masses.
(6 of them)Materials> 1 clamp stand> Stopwatch> PlasticineMethod1. Place a clamp stand on the table2. Attach the spring to the clamp stand.3. Add heavy books on the clamp stand to the base so that it is stable and doesn’t affect our readings.4.
Add Plasticine to the top of the spring so that the spring itself doesn’t vibrate a lot. Hence there would be no discrepancies in the readings.5. Add 100 grams mass to the spring.
6. Displace the 100 gram mass.7.
Release the 100 gram mass and at the same time start the stopwatch.8. Record the time taken for 10 oscillations.
9. Repeat steps 6 to 8 2 more times to get three total trials.10. Now add more mass to the spring and repeat steps 6 to 8 with 200 g, 300g, 400g, and 500g.DiagramRaw Data TableMass (grams)Time taken for 10 Oscillations (seconds)( +/- 0.
01 secs)Trial 1Trial 2Trial 31006.686.666.702009.669.
9114.99Processed Data TableFirst as we have three trials I would find the average of the three trials to get one reading for time for 10 oscillations per mass. To find the average I would add the three trials and then divide by three.Mass (grams)Time taken for 10 Oscillations (seconds)( +/- 0.01 secs)Average Time for 10 oscillations. (s)Trial 1Trial 2Trial 31006.
9914.97Now we shall find the time period. Time period is equal to the time taken for one oscillation. We have time taken for 10 oscillations, thus to find the time taken for one oscillation you simply divide the time by 10.Mass (grams)Average Time for 10 oscillations. (s)Time Period (seconds)1006.680.
971.497Now we shall plot this information on a graph.GraphThis is a graph of the time period of the spring against the independent mass suspended from it.
It clearly shows a linear positive co-relation. Linear positive co-relation means that both of the quantities are directly proportionate. We can clearly see this in the graph that as the mass suspended from the spring increases the time period of the spring increases as well. There might be only one anomaly in this graph at 200 grams.
It is a little off and thus distorts the line of best fit a little bit.ConclusionWe can prove this as the mass increasing is the same as force increasing. Thus if we increase the force the extension of the spring will also increase.
And the spring constant remains a constant value. Now if the extension of the spring increases, the amplitude increases, eventually increasing the time period. Thus we can conclude that the force or mass suspended from a spring is directly proportionate to its time period and our hypothesis is proved correct. As mass increases the time period increases as well. This can also be related to Hookes Law.
If we find the gradient of the line we should approximately get the spring constant (k) of the spring.Evaluation* The spring was old and had been used many times thus the spring did not always oscillate vertically. It moved from side to side. To avoid this we should have made sure that the it was released straight.* I could have used 2 different springs for better results.* I could have also noted down the extension in this experiment hence proving Hookes Law as well.* There might have been a human error in stopping the stopwatch exactly after 10 oscillations.Modifications* I used Plasticine in my experiment to keep the spring more stable.* A also put a heavy weight on the clamp stand so that it does not let anything vibrate which might have altered our results.Fair Test* I kept the same displacement for each trial.* I measured it for the same number of oscillations.Safe Test* We should have handled the mass with more care because the suspended mass was dangerous and kept falling and could break as well.