Physics Asked on October 3, 2021
Lets say that we have a spring, with spring constant k. It has an orignal length of $l_1$ and after stretching it a length of $l_2$. Now lets take and element of length of $Delta$l.When the spring was stretched I wanted to find out how much does this element stretch. Online I found a answer which said that as all elements stretch and equal length. $frac{Delta l }{Delta y}$=$frac{l_1}{l_2}$. Now I did not really get why this was happening. I would like a answer with a mathematical approach
Okay we have quantity called Youngs Modulus.
Let us take an elementary part of the spring with original length dl.Let us apply a force F on the spring which causes an expansion dx .Let the cross sectional area be A.Then Young Modulus is
Y = Stress / Strain = (F/A)/ (dx/dl)
Now Youngs modulus is same for all parts of the spring as it depends on material which is same for whole spring . So the quantity on the right must also be same for all elements.
Notice the above diagram. Notice how equal and opposite forces act on the ends.This is true for any dl element . This means that equal force act on each element.This is based on Newtons third law considering any element dl as our body.Further cross section are is same.
Therefore dx/dl must be same for all elements . Similar arguments may be made for whole spring.
Therefore , dx/dl = total elongation / total length .
Notice these arguments are valid as long as spring is uniform.
P.S. This can be understood using symmetry too.Each element must elongate equally as all elements are essentially same and we do not have any reason to give special treatment to only a few elements.
Correct answer by Tony Stark on October 3, 2021
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