*Parallel 0.045 * 47 = 2.12 we have closest force to what we originally added.Evaluation Many errors could have occurred during this process.*

*Parallel 0.045 * 47 = 2.12 we have closest force to what we originally added.*

You could then confirm this, and at the same time show that your results with the 2 springs, both series and parallel, are consistent with this.

Even if it's not possible to take these measurements, you should still be able to conclude that the springs were not of equal stiffness.

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?

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.

So from Hooke's law you can get x1 = F / k1 and x2 = F / k2.

Knowing this, gives me a new formula: F = k (F / k1 F / k2).

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.

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