Answer to Question #77631 in Physics for Muhammad Zeeshan

Question

Two springs A and B initially at rest have the same length L = 6 cm. When the same mass is hung from them the length of A becomes equal to 10 cm and that of B to 8 cm. What can we say about the elastic constants of the two springs?
KA = 4/5 KB
KB = 1/2 KA
KB = 4/5 KA
KB = KA
KA = 1/2 KB

Answer 1

The Hooke’s law states
F=kx
The equilibrium conditions
F_A=mg
F_B=mg
So
F_A=F_B
K_A x_A=K_B x_B
K_A/K_B =x_B/x_A =(8-6)/(10-6)=2/4=1/2
Therefore
K_A=1/2 K_B
Answer: K_A=1/2 K_B