Answer:
The effective force constant is 1233.33 N/m.
Explanation:
It is given that,
Force constant 1, [tex]k_1=3.7\times 10^3\ N/m[/tex]
Force constant 2, [tex]k_2=3.7\times 10^3\ N/m[/tex]
Force constant 3, [tex]k_3=3.7\times 10^3\ N/m[/tex]
The effective force constant if one is hung from the other in series is given by :
[tex]\dfrac{1}{K_{eff}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}+\dfrac{1}{k_3}[/tex]
[tex]\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}[/tex]
[tex]K_{eff}=1233.33\ N/m[/tex]
So, the effective force constant is 1233.33 N/m. Hence, this is the required solution.
