Also read your specification as to what they expect from conclusions, it varies arbitrarily between exam boards To add to the very sound advice above (#2), I would also suggest, if it's possible, that you go back and measure the extension of the 2nd spring separately.From the results it does look as though the springs were not identical.If I ignore the mass of the 2nd spring, the two tensions have to be equal and both equal to the force.
Also read your specification as to what they expect from conclusions, it varies arbitrarily between exam boards To add to the very sound advice above (#2), I would also suggest, if it's possible, that you go back and measure the extension of the 2nd spring separately.From the results it does look as though the springs were not identical.Tags: Creative Problem Solving GroupBobo Experiment EssaysRosenberg Research PaperArgumentative Essay On The IliadConflict In Interpersonal Relationships EssaysThesis Genetic Algorithm 2008Business Plan For TourismTeaching Through Problem Solving
The theory is based on Hooke's law which is: F = kx where F = Force, k = Constant and x = Extension [Ref. Unfortunately with the springs I have, I can only measure extension, not compression for which Hooke's is also valid.Parallel Springs: In the diagram above, T1 is the tension in spring 1, and T2 is the tension is spring 2. I assume the two springs are originally the same length and the extensions are the same in both springs. So by knowing this, you get the formula: F = k1x k2x = x (k1 k2).So overall, the spring constant for the two springs is k1 k2.Hypothesis The change in length of spring is directly proportional to the applied so that it will cause greater change in length of the spring for greater force applied.It is supported by the formula of force, F = kx, where F is the applied force, k is the spring constant of the spring, and x is the change in length or extension of the spring.I'm really stuck on writing the conclusion for this experiment.I thought I had it last night tonight it doesn't make sense any more.When we do the maths for double the mass 0.025 * 364 In series 0.011 * 202 = 2.22 The force is just over what would be expected.This is by about the same amount as a single spring.Systematic errors like measurements can only be or – 1mm certain. I'm going to finish writing up my youngs split experiment because apparently I can understand that one.Clearly two identical springs in parallel have twice as large spring constant right?