Ignore the weight of the springs.
Two springs with equal spring constants k are connected first in series (one after the other) and then in parallel (side by side) with a weight hanging from the bottom of the combination.
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What is the effective spring constant of the two different arrangements? In other words, what would be the spring constant of a single spring that would behave exactly as
(a) the series combination [Hint: Each spring stretches an amount x = F/k, but only one spring exerts a force on the hanging object.]
Answers
Answer:
As per the question, the two springs are attached to one another first in series and then in in parallel combinations.
Given : Let us consider the spring constants to be K1 and K2
To find : equivalent spring constants in series and parallel combinations
Series combinations:
in series combination we have to find out the equivalent resistance by using the reciprocal rule formula.
1/K eq. = 1/K1 + 1/K2
=> K eq. = (K1*K2)/(K1+K2)
so it would be a similar case if a single spring of spring constant K eq. be used.
Parallel combination:
in parallel combination the effective spring constant is the sum of the individual spring constants of the combination.
K eq. = (K1 + K2)
So a single spring having a spring constant of (K1+ K2) would have given us the same result.