Question

The given setup is in equilibrium springs were in natural length before force was applied. If I have to replace the above setup by 1 block and 1 spring then what will be the spring constant of that new spring, such that the block is displaced by the same amount previously by the applied force.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Assuming spring 1 is elongated by $x_1$. Spring 2 will be elongated by lets say $x_2$.

Now since, spring 2 is connected to spring 1 so free body diagram of connecting point

$K_1{X}_1\leftarrow \bullet \to K_2{X}_2$

$K_1{X}_1=K_2{X}_2=F$ .................(i)

$F=K_0{X}_0$ .................(ii)

Now if I replace this with 1 spring

Now the displacement of the block has to be same.

This means $x_1+x_2=x_0$ ...................(iii)

From eq. (1)

$x_1=F/K_1$

$x_2=F/K_2$

From eq., (ii)

$x_0=F/K_0$

Substituting in eq (iii)

$\frac{1}{K_0}=\frac{1}{K_1}+\frac{1}{K_2}$