The length of a spring is l and its spring constant is k .it is cut into two parts of length l1 and l2 and l1=nl2 .the spring constant k1 of part l2 will be
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Considering the spring to be ideal of length L, if it is cut into 1:2 ratio, the spring constant will get changed to 3k and 1.5k(3k/2) for L/3 and 2L/3 part of it, respectively.
Spring constant is not a universal constant. It depends on spring diameter, no. of windings and elastic modulus. Reducing the length of a spring reduces the number of windings on which spring constant is inversely dependent. Thus, it increases the spring constant for a spring having smaller length (small number of windings).
Spring constant is not a universal constant. It depends on spring diameter, no. of windings and elastic modulus. Reducing the length of a spring reduces the number of windings on which spring constant is inversely dependent. Thus, it increases the spring constant for a spring having smaller length (small number of windings).
grishikaagarwal03:
I can't understand
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K-1= (l1/l1+l2) k
K2=(l2/l1+l2) k
K2=(l2/l1+l2) k
Can u explain me
K1 - spring constant of length l1
K2- spring constant of length l2
But its answer is only in terms of l1
you substitute l2 = l1/n
as it is in the sum
Ohk thnx
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