Question

Suppose you have three springs with force constants of k1 = k2 = k3 = 3.70 x 10^3 N/m. What is their effective force constant if one is hung from the other in series? You may assume the springs have negligible mass.
The answer should be in N/m.

Answer 1

Answer:

The effective force constant is 1233.33 N/m.

Explanation:

It is given that,

Force constant 1, $k_1=3.7\times 10^3\ N/m$

Force constant 2, $k_2=3.7\times 10^3\ N/m$

Force constant 3, $k_3=3.7\times 10^3\ N/m$

The effective force constant if one is hung from the other in series is given by :

$\dfrac{1}{K_{eff}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}+\dfrac{1}{k_3}$

$\dfrac{1}{K_{eff}}=\dfrac{1}{3.7\times 10^3}+\dfrac{1}{3.7\times 10^3}+\dfrac{1}{3.7\times 10^3}$

$K_{eff}=1233.33\ N/m$

So, the effective force constant is 1233.33 N/m. Hence, this is the required solution.